Solve using the quadratic formula

This lesson shows how to solve using the quadratic formula. To use the quadratic formula, you need to identify a, b, and c using the standard form of a quadratic equation ax² + bx + c = 0.

First, study carefully the example in the image below. If you do not quite understand it do not worry. Keep reading to see more examples.

How to identify a, b, and c

The standard form is ax² + bx + c = 0. 

1) for 6x² + 8x + 7 = 0, we get a = 6, b = 8, and c = 7

2) for x² + 8x - 7 = 0, we get a = 1, b = 8, and c = -7

3) for -x² - 8x + 7 = 0, we get a = -1, b = -8, and c = 7

More examples showing how to solve using the quadratic formula

Example #1:

Solve using the quadratic formula x² + 8x + 7 = 0

a = 1, b = 8, and c = 7

x = [-b ± √(b² - 4ac))] / 2a

x = [-8 ± √(8² - 4 × 1 × 7)] / 2 × 1

x = [-8 ± √(64 - 4 × 1 × 7)] / 2

x = [-8 ± √(64 - 4 × 7)] / 2

x = [-8 ± √(64 - 28)] / 2

x = [-8 ± √(36)] / 2

x = (-8 ± 6 ) / 2

x₁ = (-8 + 6 ) / 2

x₁ = (-2 ) / 2

x₁ = -1

x₂ = (-8 - 6 ) / 2

x₂ = (-14 ) / 2

x₂ = -7


Example #2:

Solve using the quadratic formula 4x² - 11x - 3 = 0

a = 4, b = -11, and c = -3

x = [-b ± √(b² - 4ac)] / 2a

x = [- -11 ± √( (-11)²  - 4 × 4 × -3)] / 2 × 4

x = [11 ± √(121 - 4 × 4 × -3)] / 8

x = [11 ± √(121 - 4 × -12)] / 8

x = [11 ± √(121 + 48)] / 8

x = [11 ± √(169)] / 8

x = (11 ± 13 ) / 8

x₁ = (11 + 13 ) / 8

x₁ = (24 ) / 8

x₁ = 3

x₂ = (11 - 13 ) / 8

x₂ = (-2 ) / 8

x₂ = -1/4

Example #3:

Solve using the quadratic formula x² + x - 2 = 0

a = 1, b = 1, and c = -2

x = [-b ± √(b² - 4ac)] / 2a

x = [-1 ± √((1)²  - 4 × 1 × -2)] / 2 × 1

x = [-1 ± √(1 - 4 × 1 × -2)] / 2

x = [-1 ± √(1 - 4 × -2)] / 2

x = [-1 ± √(1 + 8)] / 2

x = [-1 ± √(9)] / 2

x = (-1 ± 3 ) / 2

x₁ = (-1 + 3 ) / 2

x₁ = (2) / 2

x₁ = 1

x₂ = (-1 - 3 ) / 2

x₂ = (-4 ) / 2

x₂ = -2

Notice that in all three examples above, the discriminant is positive and the number of solutions of each quadratic equation is 2. For example, in example #3, the discriminant shown in blue is 9 and the solutions are 1 and -2.

Example #4:

Solve using the quadratic formula 4x² + 12x + 9 = 0

a = 4, b = 12, and c = 9

x = [-b ± √(b² - 4ac)] / 2a

x = [-12 ± √((12)²  - 4 × 4 × 9)] / 2 × 4

x = [-12 ± √(144 - 4 × 4 × 9)] / 8

x = [-12 ± √(144 - 4 × 36)] / 8

x = [-12 ± √(144 - 144)] / 8

x = [-12 ± √(0)] / 8

x = (-12 ± 0) / 8

x₁ = (-12 + 0) / 8

x₁ = (-12) / 8

x₁ = -1.5

x₂ = (-12 - 0) / 8

x₂ = (-12) / 8

x₂ = -1.5

Notice that in example #4, the discriminant shown in blue is zero and there is only one solution. 

Example #5:

Solve using the quadratic formula x² + 8x + 25 = 0

a = 1, b = 8, and c = 25

x = [-b ± √(b² - 4ac)] / 2a

x = [-8 ± √((-8)²  - 4 × 1 × 25)] / 2 × 1

x = [-8 ± √(64 - 4 × 25)] / 2

x = [-8 ± √(64 - 100)] / 2

x = [-8 ± √(-36)] / 2

x = [-8 ± √(36(-1))] / 2

x = [-8 ± √(36(i²))] / 2

x = (-8 ± 6i) / 2

x₁ = (-8 + 6i) / 2

x₁ = -8/2 + 6i/2

x₁ = -4 + 3i

x₂ = (-8 - 6i) / 2

x₂ = -8/2 - 6i/2

x₂ = -4 - 3i

Notice that in example #5, the discriminant shown in blue is negative and you ended up with two complex numbers. Since it is not possible to take the square root of a negative number, mathematicians came up with a solution by letting i² equal to -1 and i is called imaginary number.

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