Mathematics · Algebra

Quadratic formula

Q: How do you solve a quadratic equation with the quadratic formula?

First bring the equation into the form $ax^2+bx+c=0$ and read off a, b and c. Then insert them into $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. First compute the discriminant $b^2-4ac$ under the root, take the root and form the two solutions with the plus and the minus sign. Example: $2x^2-4x-6=0$ gives $x=\frac{4\pm\sqrt{16+48}}4=\frac{4\pm8}4$, that is $x_1=3$ and $x_2=-1$. Watch the signs of b and c and do not forget the ±, otherwise you lose one solution. For real-root evaluation, require $a,b,c\in\mathbb R$, $a\ne0$ and $D\ge0$. Finally substitute each candidate into the original equation to check the result.

Q: What does the discriminant tell you about the solutions?

The discriminant $D=b^2-4ac$ sits under the root and decides the number and type of real solutions. If D is greater than zero, two distinct real solutions exist, because the root gives a positive value that is once added and once subtracted. If D equals zero, both solutions merge into a single double root $x=-\frac b{2a}$. If D is less than zero, there is no real solution, only two complex ones, because you cannot take a real root of a negative number. So from the sign of the discriminant alone you can judge the solvability before calculating further.

Q: What is the difference between the quadratic formula and the pq formula?

Both formulas solve quadratic equations but assume different starting forms. The quadratic or ABC formula works with the general form $ax^2+bx+c=0$ and gives $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$; the leading factor a may be any nonzero real number. The pq formula, by contrast, requires the normalized form $x^2+px+q=0$ with leading coefficient one and gives $x=-\frac p2\pm\sqrt{\left(\frac p2\right)^2-q}$. If you want to use the pq formula, you first divide the equation by a. Both lead to the same result; the choice is a matter of taste and often depends on which form the problem provides.

Q: What does it mean when the discriminant is exactly zero?

If the discriminant $b^2-4ac$ equals zero, the root vanishes and the two solutions of the formula merge into a single one: $x=-\frac b{2a}$. This is called a double root. Geometrically it means that the associated parabola does not cross the x-axis but only touches it; the vertex lies exactly on the axis. This special situation occurs in problems about tangency of curves or in limiting cases, for example when a line just touches a parabola. If you recognize $D=0$, you no longer need the ±, because both signs give the same result.

Q: How are the solutions related to Vieta formulas?

Vieta formulas link the two solutions x₁ and x₂ of a quadratic equation $x^2+px+q=0$ directly with the coefficients: the sum of the solutions is $x_1+x_2=-p$, their product is $x_1x_2=q$. With this you can often find solutions in your head or check a result without going through the whole formula. If you find two numbers whose sum is −p and whose product is q, those are exactly the roots. Example: $x^2-5x+6=0$ has solutions with sum 5 and product 6, that is 2 and 3. Vieta is excellent for quickly checking the values you computed with the formula.

The quadratic formula determines roots of $ax^2+bx+c=0$ , with $a\ne0$ . The discriminant tells whether there are two, one or no real roots.

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Formula

Calculate

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Dimensionless (algebra)

Variables & units

Variables & units
Symbol	Meaning	Unit
$a$	Coefficient of $x^{2} (a \ne{} 0)$	dimensionless
$b$	Coefficient of x	dimensionless
$c$	Constant term	dimensionless
$D$	Discriminant $D = b^{2} - 4ac$	dimensionless

LaTeX

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

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Diagram: an upward-opening parabola crosses the x-axis at the two roots x₁ and x₂, the vertex lies below the axis. — Parabola with two roots x₁ and x₂, the two solutions of the quadratic equation (D > 0).

Quadratic formula calculator

Find both real solutions of ax² + bx + c = 0 and the discriminant, with a check; for a = 0 the linear equation.

Calculation model

Find

Coefficient a

Number in front of x² in standard form, not 0, e.g. 2.

1

Coefficient b

Number in front of x with its sign, e.g. −4.

1

Coefficient c

Number without x, with its sign, e.g. −6.

1

Filled with example values. Type your own and the result updates straight away.

Result: Smaller real root x₁

x₁ = -1

With a = 2, b = -4 and c = -6, x₁ = -1.

Working · rearrange for x₁, substitute, evaluate

1. Rearrange for x₁
$\begin{aligned}D&=b^2-4ac\\x_{1,2}&=\frac{-b\pm\sqrt D}{2a}\quad(D\ge0)\end{aligned}$
2. Substitute values (in SI units)
$\begin{aligned}D&=(-4)^2-4\cdot2\cdot(-6)=64\\x_{1,2}&=\frac{-(-4)\pm\sqrt{64}}{2\cdot2}\\x_1&=-1\\x_2&=3\\a\,x_1^2+b\,x_1+c&=2\cdot(-1)^2+(-4)\cdot(-1)+(-6)=0\\a\,x_2^2+b\,x_2+c&=2\cdot3^2+(-4)\cdot3+(-6)=0\end{aligned}$
3. Result
x₁ = -1

D > 0: two distinct real roots in ascending order (x₁ < x₂).

Then check the original domain and the conditions of an applied problem. A satisfied check shows a root of the equation, not automatically its physical admissibility.

Quick tips

First bring the equation into standard form, then read off the coefficients with their signs: 3x² + 2 = 7x becomes 3x² − 7x + 2 = 0, so b = −7.
The discriminant D = b² − 4ac tells you what to expect: D > 0 two solutions, D = 0 one repeated solution, D < 0 no real one.

Assumptions and limits

Real coefficients in established standard form ax²+bx+c=0, a≠0. Coefficients are evaluated exactly as decimals, beyond 15 significant digits as the nearest floating-point value. The sign and zero case of the discriminant are therefore exact: D = 0 gives exactly one repeated root, a negative discriminant no real root. Roots use the numerically stable q formulation and appear in the working with a check. Complex numbers remain a separate article explanation.

Derivation & background

Completing the square transforms the equation into a perfect square. For real coefficients, $D=b^2-4ac$ determines the roots: positive gives two distinct real roots, zero a repeated root, and negative no real roots. Complex roots require a different number system.

Exam blueprint

Validity range

Applies to $ax^2+bx+c=0$ with real coefficients and $a\ne0$ . Real-root evaluation requires $D\ge0$ ; complex roots are explained separately.

Derivation steps

Completing the square converts $ax^2+bx+c=0$ into a perfect square; both square-root branches give the solution set.

1Establish $ax^2+bx+c=0$ and check $a\ne0$ .
2Divide by a and subtract $\frac ca$ : $x^2+\frac ba x=-\frac ca$ .
3Add $\left(\frac{b}{2a}\right)^2$ to both sides: $\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$ .
4Calculate $D=b^2-4ac$ . For $D\ge0$ take both square-root signs; note $\sqrt{4a^2}=2|a|$ .
5Subtract $\frac{b}{2a}$ : $x=\frac{-b\pm\sqrt D}{2a}$ . A negative a merely swaps the order of the two branches.
6 $D<0$ means no real roots. Check roots in the original equation and, where relevant, against application constraints.

Rearrangements

Determine discriminant

D = b^2 - 4ac

D determines number and type of real solutions.

Task variant

What does $D=0$ mean?

There is exactly one real double root $x=-\frac b{2a}$ .

Common mistakes

Forgetting the ± before the square root.

For $D>0$ there are two solutions.

Exam context

Typical for roots, projectile parabolas, intersections and optimization.

Formula cluster

Algebraic solving

Connects equations, parabolas and discriminant analysis.

a² + b² = c²

Pythagorean Theorem

d/dx(xⁿ) = n·xⁿ⁻¹

Power Rule of Differentiation

f(x) = Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ

Taylor Series

Worked example

$2x^2-4x-6=0$ → $x=\frac{4\pm\sqrt{16+48}}4=\frac{4\pm8}4$ → $x_1=3$ , $x_2=-1$ .

Applications

Quadratic roots, intersections, geometric area problems and motion under constant acceleration

How do you calculate with Quadratic Formula?

Here is how to work through a typical Quadratic Formula (x = (−b ± √D) / 2a) task step by step:

1
Task
What does $D=0$ mean?
Solution path
There is exactly one real double root $x=-\frac b{2a}$ .

Understand and apply the equation

The quadratic formula solves quadratic equations even when their factors are not obvious. This guide explains each transformation, from standard form and completing the square to the discriminant, substitution checks and choosing a meaningful solution in an application.

Understand what you are solving

You are looking for numbers that make both sides of an equation equal. A quadratic equation also contains the unknown squared. Two numbers may work, one may be a repeated root, or real roots may be absent. The quadratic formula gives candidate values; substitution checks whether they satisfy the original equation.[1]

Prerequisites: collect like terms, perform the same operation on both sides and square negative numbers in parentheses.
Specify the number system: many school problems ask for real roots. Complex roots are introduced in the labelled extension.

Start by establishing standard form

The general form is $ax^2+bx+c=0$ , with $a,b,c\in\mathbb R$ and $a\ne0$ . Move all terms to one side and collect like powers. A missing linear term gives $b=0$ ; a missing constant term gives $c=0$ . Always include the sign when reading a coefficient.[1]

x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

The plus sign gives one formula branch and the minus sign the other.

If $a=0$ , the equation is not quadratic: when $b\ne0$ , x = $-\frac cb$ . If b and c are also zero, every x satisfies it; if $b=0$ and $c\ne0$ , no x does. Division by 2a is invalid in all of these cases.

Derive the formula by completing the square

We seek a perfect square. Since $a\ne0$ , division by a is allowed. Then move $\frac ca$ to the right-hand side.[2]

ax^2+bx+c=0\quad\Longleftrightarrow\quad x^2+\frac ba x=-\frac ca

Divide by a and subtract $\frac ca$ .

The identity $(x+d)^2=x^2+2dx+d^2$ identifies the missing term. Choose $d=\frac{b}{2a}$ to match the middle term, then add $d^2$ to both sides. The left side becomes a square while the solution set stays unchanged.

\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}-\frac ca=\frac{b^2-4ac}{4a^2}

Complete the square on the left and combine the right-hand terms over a common denominator.

For $D=b^2-4ac\ge0$ , taking square roots gives both signs. Strictly, $\sqrt{4a^2}=2|a|$ . Replacing |a| with a merely exchanges the two branches when a is negative, leaving the solution set unchanged.

x=-\frac{b}{2a}\pm\frac{\sqrt D}{2|a|}\quad\Longrightarrow\quad x_{1,2}=\frac{-b\pm\sqrt D}{2a}

Subtract $\frac{b}{2a}$ and express both solution branches together.

What the discriminant and graph reveal

$D>0$ : two distinct real solutions; the parabola crosses the x-axis twice.
$D=0$ : a repeated real root x = $-\frac{b}{2a}$ ; the parabola touches the x-axis at its vertex.
$D<0$ : no real solutions. Over the complex numbers, real coefficients give two conjugate roots.

[1][3]

f(x)=a\left(x+\frac{b}{2a}\right)^2-\frac{D}{4a},\qquad S=\left(-\frac{b}{2a},-\frac{D}{4a}\right)

Vertex form and vertex coordinates of the associated quadratic function.[3]

The sign of a determines opening direction: upward if positive, downward if negative. D determines the number of real roots. These are separate facts: a downward-opening parabola may also have two, one or no real roots.[3]

Two crossings, one tangency or no intersection

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The original functions

y=(x-2)^2-1

,

y=(x-2)^2

and

y=(x-2)^2+1

have discriminants 4, 0 and −4. Their real roots are 1 and 3, only 2, and none, respectively. Both axes show dimensionless numbers. Use the legend, vertical position and solid-versus-dashed lines to identify the cases; both dashed curves use the same line pattern.[3][1]

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The vertex form of a parabola relates the discriminant to the vertex position: its y-coordinate is $-\frac{D}{4a}$ . The sign of a alone determines whether the parabola opens upward or downward; the discriminant does not decide this.[3]

Three examples with an approach and checks

Handle negative coefficients correctly

Given: $3x^2+2=7x$ . Find all real solutions.
Find: Establish standard form, evaluate both branches and check both roots.

Approach: Put the equation into standard form first so the linear coefficient has an unambiguous sign. The discriminant determines real branches; check both roots in the original equation.
$\begin{aligned}ax^2+bx+c&=0\\{}D&=b^2-4ac\\{}x_{1,2}&=\frac{-b\pm\sqrt D}{2a}\end{aligned}$
Subtract 7x: $3x^2-7x+2=0$ . Thus $a=3,\ b=-7,\ c=2$ .
Square b using parentheses and subtract 4ac.
$D=(-7)^2-4\cdot3\cdot2=49-24=25$
Substitute $-b=7$ and $\sqrt D=5$ ; divide the entire numerator by 6.
$x_{1,2}=\frac{7\pm5}{6}\quad\Longrightarrow\quad x_1=2,\;x_2=\frac13$
Substitute into the original equation: for $x=2$ both sides equal 14; for $x=\frac13$ both sides equal $\frac73$ .

Result: The solution set is $\left\{\frac13,2\right\}$ . Both distinct real roots have been checked.

Check: Additional check: the sum is $\frac73=-\frac ba$ and the product $\frac23=\frac ca$ . Naming either root $x_1$ or $x_2$ does not change the solution set.

Choose a meaningful root in an application

Given: A rectangle has area 84 cm². Its length exceeds its width by 5 cm. x is the numerical width in centimetres.
Find: Find width and length.

Approach: Numerical width in centimetres is x, so length is x plus five. The area condition gives a quadratic equation; only a positive width is admissible in the physical model.
$\begin{aligned}x(x+5)&=84\\{}x^2+5x-84&=0\end{aligned}$
Write area as a product: $x(x+5)=84$ . Standard form: $x^2+5x-84=0$ .
Calculate the discriminant: $25+336=361=19^2$ .
$x_{1,2}=\frac{-5\pm19}{2}\quad\Longrightarrow\quad x_1=7,\;x_2=-12$
Width must be positive. $x=-12$ is algebraically correct but does not describe this rectangle.

Result: Width 7 cm, length 12 cm.

Check: $7\cdot12=84$ and $12-7=5$ . Check application constraints after solving; negative roots are not invalid for quadratic equations in general.

Why there is no real root

Given: $2x^2-4x+7=0$ ; find its real roots.
Find: Explain the absence of real intersections algebraically and geometrically.

Approach: the discriminant determines whether the square root is real. The vertex form gives an independent check.
$D=b^2-4ac$
Substitute the signed coefficients.
$D=(-4)^2-4\cdot2\cdot7=16-56=-40<0$
A real square is never negative. In vertex form, the entire expression on the left is therefore always positive.
$2x^2-4x+7=2(x-1)^2+5\ge5$

Result: There are no real roots.

Check: The parabola opens upward and has minimum value five. It never reaches the x-axis.

Quadratic formula, normalised formula or factoring?

The quadratic formula works for every $a\ne0$ . The normalised p–q formula first requires division by a: p = $\frac ba$ and q = $\frac ca$ . Obvious factors may make the zero-product property quicker. For $x^2=d$ , taking both square-root signs is sufficient when $d\ge0$ .[2]

x_{1,2}=-\frac p2\pm\sqrt{\left(\frac p2\right)^2-q},\qquad p=\frac ba,\;q=\frac ca

The p–q formula for the normalised equation $x^2+px+q=0$ .

All valid methods must produce the same solution set. In Example 1, $(3x-1)(x-2)=0$ independently checks both values. If the original problem involved a denominator or an earlier squaring step, also check the original domain and equation.

The p–q formula is equivalent after correct normalisation. Preserve signs during division: divide b and c by the same nonzero a. Then choose the approach whose intermediate steps you can check confidently.[1][2]

Explore further: complex roots

A negative discriminant rules out real roots, but not solutions in every number system. In the complex numbers, $i^2=-1$ . Real coefficients and $D<0$ give a conjugate pair. Use this extension only when the requested number system permits it.[1]

x_{1,2}=\frac{-b\pm i\sqrt{-D}}{2a},\qquad a,b,c\in\mathbb R,\;a\ne0,\;D<0

Here −D is positive. The two values have opposite imaginary parts.

Five exercises with solutions

Calculate · Foundation

1. Solve $2x^2+5x-3=0$ over the real numbers.

Show hint

$a=2,\ b=5,\ c=-3$ . Calculate D first.

Learning goal: Find two real roots and check them.

Show solution and calculation

Approach: Calculate $D=b^2-4ac$ ; for a nonnegative result use $x_{1,2}=\frac{-b\pm\sqrt D}{2a}$ .
Here $D=25+24=49$ .
$x=\frac{-5\pm7}{4}$ gives $\frac12$ and $-3$ .
Checks: $2\left(\frac12\right)^2+5\left(\frac12\right)-3=0$ ; $18-15-3=0$ .

x_{1,2}=\frac{-5\pm7}{4}

Calculate · Application

2. Solve $x^2-6x+9=0$ . How many distinct roots are there?

Show hint

Compare the discriminant with zero.

Learning goal: Recognize a repeated root.

Show solution and calculation

Approach: Determine $D=b^2-4ac$ ; if $D=0$ , the repeated root is $x=-\frac{b}{2a}$ .
Here $D=36-36=0$ .
There is one distinct root $x=3$ with multiplicity two.
Factoring confirms $(x-3)^2=0$ .

Calculate · Application

3. Solve $x^2+4x+8=0$ first over ℝ, then over ℂ.

Show hint

D is negative. For complex roots, use $i^2=-1$ .

Learning goal: Distinguish real and complex roots.

Show solution and calculation

Approach: Check $D=b^2-4ac$ .
For $D<0$ there are no real roots; over $\mathbb C$ , use $x_{1,2}=\frac{-b\pm i\sqrt{-D}}{2a}$ .
Here $D=16-32=-16$ .
There are no real roots; the complex roots are $x=\frac{-4\pm4i}{2}=-2\pm2i$ .

Check: $(x+2)^2+4=0$ , since $(\pm2i)^2=-4$ .

x_{1,2}=-2\pm2i

Interpret · Application

The parabola $y=x^2-4x+k$ has an unknown real parameter k. Which k values give two, one or no distinct real roots? Give the contact point in the tangent case.

Show hint

The parameter shifts the parabola vertically; use the discriminant.

Learning goal: Determine tangential axis contact from a parameter.

Show solution and calculation

Approach: The discriminant is sixteen minus four k.
Positive values give two crossings for k below four; zero gives tangency at k equal to four; negative values give no crossing.
Completing the square places the contact point at (2,0).
It is one distinct root with multiplicity two.

\begin{aligned}D&=16-4k\\{}y&=(x-2)^2+(k-4)\\{}k=4&\Rightarrow x=2\end{aligned}

[1]

Error analysis · Application

For $x^2-5x=0$ someone divides by x and obtains only $x=5$ . Check the step and recover the complete real solution set with substitution.

Show hint

Division requires a nonzero divisor.

Learning goal: Check division by an unknown for lost solutions.

Show solution and calculation

Approach: Factor without dividing by the unknown.
The product x times (x−5) is zero precisely for x equal to zero or five.
Both solve the original equation.
Division was valid only in the nonzero case and discarded the other case; the quadratic formula gives the same two roots.

\begin{aligned}x(x-5)&=0\\{}x_{1,2}&=\frac{5\pm\sqrt{25}}2\in\{0,5\}\end{aligned}

[1]

Avoid errors and interpret the result

Do not read coefficients before establishing standard form: in $3x^2+2=7x$ , b is −7, not +7.
Square negative numbers in parentheses: $(-7)^2=49$ , whereas $-7^2=-49$ .
Divide the entire numerator by 2a. $-b+\frac{\sqrt D}{2a}$ is a different expression.
Avoid premature rounding. Keep exact fractions or radicals until the final result.
Evaluate both signs and substitute every root into the original equation.

The formula solves quadratic equations. A fourth-degree equation can use it only when a valid substitution produces a quadratic; afterwards, reverse the substitution and check its constraints. The formula alone does not perform that check.

What to remember when solving

First put the equation into general form with zero on one side. Record the three signed coefficients and calculate the discriminant. Evaluate real square-root branches only when the discriminant is nonnegative. Check every result in the original equation and enforce any application-specific restrictions.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 2, 2026Added relevant formula connections and model distinctions; preserved existing derivations, exercises and original figures.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Clarified the introduction and prerequisites; added a third worked example, an original figure and a summary; made supplemental material selectively expandable.

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Frequently asked questions about Quadratic Formula

How do you solve a quadratic equation with the quadratic formula?+

First bring the equation into the form $ax^2+bx+c=0$ and read off a, b and c. Then insert them into $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . First compute the discriminant $b^2-4ac$ under the root, take the root and form the two solutions with the plus and the minus sign. Example: $2x^2-4x-6=0$ gives $x=\frac{4\pm\sqrt{16+48}}4=\frac{4\pm8}4$ , that is $x_1=3$ and $x_2=-1$ . Watch the signs of b and c and do not forget the ±, otherwise you lose one solution. For real-root evaluation, require $a,b,c\in\mathbb R$ , $a\ne0$ and $D\ge0$ . Finally substitute each candidate into the original equation to check the result.

What does the discriminant tell you about the solutions?+

The discriminant $D=b^2-4ac$ sits under the root and decides the number and type of real solutions. If D is greater than zero, two distinct real solutions exist, because the root gives a positive value that is once added and once subtracted. If D equals zero, both solutions merge into a single double root $x=-\frac b{2a}$ . If D is less than zero, there is no real solution, only two complex ones, because you cannot take a real root of a negative number. So from the sign of the discriminant alone you can judge the solvability before calculating further.

What is the difference between the quadratic formula and the pq formula?+

Both formulas solve quadratic equations but assume different starting forms. The quadratic or ABC formula works with the general form $ax^2+bx+c=0$ and gives $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ; the leading factor a may be any nonzero real number. The pq formula, by contrast, requires the normalized form $x^2+px+q=0$ with leading coefficient one and gives $x=-\frac p2\pm\sqrt{\left(\frac p2\right)^2-q}$ . If you want to use the pq formula, you first divide the equation by a. Both lead to the same result; the choice is a matter of taste and often depends on which form the problem provides.

What does it mean when the discriminant is exactly zero?+

If the discriminant $b^2-4ac$ equals zero, the root vanishes and the two solutions of the formula merge into a single one: $x=-\frac b{2a}$ . This is called a double root. Geometrically it means that the associated parabola does not cross the x-axis but only touches it; the vertex lies exactly on the axis. This special situation occurs in problems about tangency of curves or in limiting cases, for example when a line just touches a parabola. If you recognize $D=0$ , you no longer need the ±, because both signs give the same result.

How are the solutions related to Vieta formulas?+

Vieta formulas link the two solutions x₁ and x₂ of a quadratic equation $x^2+px+q=0$ directly with the coefficients: the sum of the solutions is $x_1+x_2=-p$ , their product is $x_1x_2=q$ . With this you can often find solutions in your head or check a result without going through the whole formula. If you find two numbers whose sum is −p and whose product is q, those are exactly the roots. Example: $x^2-5x+6=0$ has solutions with sum 5 and product 6, that is 2 and 3. Vieta is excellent for quickly checking the values you computed with the formula.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, College Algebra 2e, §2.5: Quadratic Equations

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Quadratic formula

Variables & units

Quadratic formula calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Algebraic solving

Worked example

Applications

How do you calculate with Quadratic Formula?

Understand what you are solving

Start by establishing standard form

Derive the formula by completing the square

What the discriminant and graph reveal

Two crossings, one tangency or no intersection

Three examples with an approach and checks

Handle negative coefficients correctly

Choose a meaningful root in an application

Why there is no real root

Quadratic formula, normalised formula or factoring?

Explore further: complex roots

Five exercises with solutions

1. Solve 2x2+5x−3=02x^2+5x-3=02x2+5x−3=0 over the real numbers.

2. Solve x2−6x+9=0x^2-6x+9=0x2−6x+9=0. How many distinct roots are there?

3. Solve x2+4x+8=0x^2+4x+8=0x2+4x+8=0 first over ℝ, then over ℂ.

The parabola y=x2−4x+ky=x^2-4x+ky=x2−4x+k has an unknown real parameter k. Which k values give two, one or no distinct real roots? Give the contact point in the tangent case.

For x2−5x=0x^2-5x=0x2−5x=0 someone divides by x and obtains only x=5x=5x=5. Check the step and recover the complete real solution set with substitution.

Avoid errors and interpret the result

What to remember when solving

Sources for this explanation

Frequently asked questions about Quadratic Formula

Subject references and further reading

Related formulas

More formulas: Algebra and equations

Retain Quadratic Formula for exams

1. Solve $2x^2+5x-3=0$ over the real numbers.

2. Solve $x^2-6x+9=0$ . How many distinct roots are there?

3. Solve $x^2+4x+8=0$ first over ℝ, then over ℂ.

The parabola $y=x^2-4x+k$ has an unknown real parameter k. Which k values give two, one or no distinct real roots? Give the contact point in the tangent case.

For $x^2-5x=0$ someone divides by x and obtains only $x=5$ . Check the step and recover the complete real solution set with substitution.