# 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.

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

To find x or the solution of a quadratic equation, the quadratic formula tells you to take the negative of b. Then, add or subtract negative b to the square root of b squared minus 4ac. Finally divide everything you got so far by 2a.

The plus or minus (±) in the formula means that you will end up with two answers.

You can also write the quadratic formula as x = [-b ± √(b2 - 4ac)]/2a

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

It is important to avoid substituting the wrong values for a, b, and c so you do not end up with incorrect solutions. As a math teacher, I have seen students making this mistake quite often. Therefore, study the three examples below carefully so you know 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

The value of the expression under the radical sign or b2 - 4ac has a special name. It is called discriminant! Its value will tell you the number of solutions and whether the solutions are real solutions (real roots) or complex solutions (complex roots).

## 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.

In general, when the discriminant is positive, you will get two answers.

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.

In general, when the discriminant is zero, you will get only one answer.

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.

In general, when the discriminant is negative, you will get complex numbers as solutions.

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