# Quadratic Formula

How to solve a quadratic equation by using the quadratic formula: formula, 2 examples, and their solutions.

## Formula

### Formula

For a quadratic equation

ax^{2} + bx + c = 0

(a ≠ 0),

x = [-b ± √b^{2} - 4ac] / 2a.

This is the quadratic formula.

## Example 1

### Example

### Solution

The given quadratic equation is

1x^{2} + 3x - 2 = 0.

a = 1

b = 3

c = -2

Then, by the quadratic formula,

x = [-3 ± √3^{2} - 4⋅1⋅(-2)] / [2⋅1].

3^{2} = 9

-4⋅1⋅(-2) = +8

2⋅1 = 2

9 + 8 = 17

So x = [-3 ± √17]/2.

## Example 2

### Example

### Solution

The given quadratic equation is

4x^{2} - x + 5 = 0.

a = 4

b = -1

c = +5

Then, by the quadratic formula,

x = [+1 ± √(-1)^{2} - 4⋅4⋅5] / [2⋅4].

+1 = 1

(-1)^{2} = 1

-4⋅4⋅5 = -80

2⋅4 = 8

1 - 80 = -79

See √-79.

The number in the radical sign (radicand)

is minus.

But if the radicand is minus,

it's not a real number.

So the quadratic equation has no real roots.

So [No real roots] is the answer.

To see how to write x in complex number,

click the link below.

Complex roots of a quadratic equation