Quadratic Formula Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Quadratic Formula.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

A formula giving the exact solutions to any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 directly from its three coefficients.

When factoring fails, this formula always finds the x-intercepts.

Read the full concept explanation β†’

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: The quadratic formula gives the exact solutions of ax2+bx+c=0ax^2+bx+c=0 from its three coefficients.

Common stuck point: The procedure for quadratic formula is the easy part; the trap is using the formula before setting the equation to zero. Asking "Do I have a quadratic equation set to zero whose exact solutions I need?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Do I have a quadratic equation set to zero whose exact solutions I need?

Worked Examples

Example 1

easy
Solve x2βˆ’5x+6=0x^2 - 5x + 6 = 0 by factoring.

Answer

x=2Β orΒ x=3x = 2 \text{ or } x = 3

First step

1
Find two numbers that multiply to 6 and add to βˆ’5-5: those are βˆ’2-2 and βˆ’3-3.

Full solution

  1. 2
    Factor: (xβˆ’2)(xβˆ’3)=0(x - 2)(x - 3) = 0.
  2. 3
    Set each factor to zero: xβˆ’2=0x - 2 = 0 or xβˆ’3=0x - 3 = 0.
  3. 4
    Solutions: x=2x = 2 or x=3x = 3.
Factoring works when you can find two numbers whose product equals the constant term and whose sum equals the coefficient of xx. The zero product property then gives the solutions.

Example 2

medium
Solve 2x2+3xβˆ’2=02x^2 + 3x - 2 = 0 using the quadratic formula.

Example 3

hard
Solve 2x2βˆ’5xβˆ’3=02x^2 - 5x - 3 = 0 using the quadratic formula.

Example 4

medium
Solve 3x2βˆ’4xβˆ’4=03x^2 - 4x - 4 = 0 using the quadratic formula.

Example 5

medium
A rectangular garden has area 5454 mΒ² and its length is 33 m more than its width. Find the width.

Example 6

hard
Solve 2x+3x+1=1\frac{2}{x} + \frac{3}{x + 1} = 1 for xβ‰ 0,βˆ’1x \neq 0, -1.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Solve x2βˆ’9=0x^2 - 9 = 0.

Example 2

hard
Solve x2+4x+5=0x^2 + 4x + 5 = 0 and describe the solutions.

Example 3

easy
Solve x2βˆ’5x+6=0x^2 - 5x + 6 = 0 using the quadratic formula.

Example 4

easy
Solve x2+7x+12=0x^2 + 7x + 12 = 0 using the quadratic formula.

Example 5

easy
Solve 2x2βˆ’4xβˆ’6=02x^2 - 4x - 6 = 0.

Example 6

easy
Solve x2βˆ’9=0x^2 - 9 = 0.

Example 7

easy
Solve x2+2xβˆ’8=0x^2 + 2x - 8 = 0.

Example 8

easy
Solve 3x2+2xβˆ’1=03x^2 + 2x - 1 = 0.

Example 9

easy
Solve x2βˆ’4x+4=0x^2 - 4x + 4 = 0.

Example 10

easy
Find the discriminant of 5x2βˆ’3x+1=05x^2 - 3x + 1 = 0 and state how many real roots the equation has.

Example 11

medium
Solve x2βˆ’2xβˆ’1=0x^2 - 2x - 1 = 0.

Example 12

medium
Solve 2x2+3xβˆ’5=02x^2 + 3x - 5 = 0.

Example 13

medium
Solve x2βˆ’6x+13=0x^2 - 6x + 13 = 0.

Example 14

medium
For what value of kk does x2βˆ’6x+k=0x^2 - 6x + k = 0 have exactly one real solution?

Example 15

medium
A rectangle has area 2424 and length 55 more than width. Find the dimensions.

Example 16

medium
A ball is thrown upward with initial velocity 2020 m/s from a height of 55 m. Its height is h(t)=βˆ’5t2+20t+5h(t) = -5t^2 + 20t + 5. When does it hit the ground?

Example 17

medium
Find the sum and product of the roots of 3x2βˆ’7x+2=03x^2 - 7x + 2 = 0 without solving the equation.

Example 18

medium
Solve x+6x=5x + \frac{6}{x} = 5.

Example 19

medium
Solve x4βˆ’5x2+4=0x^4 - 5x^2 + 4 = 0.

Example 20

challenge
If the roots of x2βˆ’8x+k=0x^2 - 8x + k = 0 are in the ratio 3:13 : 1, find kk.

Example 21

challenge
Find all values of aa for which x2+ax+4=0x^2 + ax + 4 = 0 has two distinct positive real roots.

Example 22

challenge
Prove that for any real aa, the equation x2+ax+(aβˆ’1)=0x^2 + ax + (a - 1) = 0 has at least one real solution.

Example 23

easy
Identify aa, bb, and cc in 4x2βˆ’7x+2=04x^2 - 7x + 2 = 0.

Example 24

easy
Solve x2βˆ’6x+8=0x^2 - 6x + 8 = 0 using the quadratic formula.

Example 25

easy
Solve x2+3xβˆ’10=0x^2 + 3x - 10 = 0.

Example 26

easy
Solve x2+6x+9=0x^2 + 6x + 9 = 0.

Example 27

medium
Solve x2βˆ’4x+1=0x^2 - 4x + 1 = 0. Leave answers in exact form.

Example 28

medium
Solve 2x2+xβˆ’6=02x^2 + x - 6 = 0.

Example 29

medium
Solve x2βˆ’2x+5=0x^2 - 2x + 5 = 0 in the complex numbers.

Example 30

medium
Find the values of kk for which x2+kx+4=0x^2 + kx + 4 = 0 has a double root.

Example 31

medium
Solve 4x2βˆ’12x+9=04x^2 - 12x + 9 = 0.

Example 32

medium
Solve x2+5xβˆ’14=0x^2 + 5x - 14 = 0.

Example 33

medium
Solve 5x2βˆ’2xβˆ’3=05x^2 - 2x - 3 = 0.

Example 34

hard
Solve 2x2+6x+1=02x^2 + 6x + 1 = 0. Leave answer in exact form.

Example 35

hard
A projectile follows h(t)=βˆ’5t2+25t+1h(t) = -5t^2 + 25t + 1 metres. When does it reach height 2020 m? Give all solutions to two decimal places.

Example 36

hard
Find values of kk such that x2+(k+2)x+9=0x^2 + (k+2)x + 9 = 0 has no real roots.

Example 37

hard
If one root of x2βˆ’7x+k=0x^2 - 7x + k = 0 is 33, find kk and the other root.

Example 38

hard
Find bb such that the equation x2+bx+16=0x^2 + bx + 16 = 0 has roots whose difference is 66.

Example 39

hard
Solve x4βˆ’10x2+9=0x^4 - 10x^2 + 9 = 0.

Example 40

hard
The roots of 2x2βˆ’5x+c=02x^2 - 5x + c = 0 differ by 11. Find cc.

Example 41

challenge
Prove that the equation x2βˆ’(a+b)x+abβˆ’1=0x^2 - (a + b)x + ab - 1 = 0 always has two distinct real roots for any real aa, bb.

Background Knowledge

These ideas may be useful before you work through the harder examples.

quadratic functionssquare roots