# Quadratic Equation: Complex Roots

How to find the complex roots of a quadratic equation and see if the quadratic equation has two complex roots: 2 formulas, 2 examples, and their solutions.

## Formula

### Formula

Recall that

for a quadratic equation

ax^{2} + bx + c = 0

(a ≠ 0),

x = [-b ± √b^{2} - 4ac] / 2a.

This is the quadratic formula.

Let's use this

to find the complex roots.

## Example 1

### Example

### Solution

The given quadratic equation is

2x^{2} - 3x + 5 = 0.

a = 2

b = -3

c = +5

Then, by the quadratic formula,

x = [+3 ± √(-3)^{2} - 4⋅2⋅5] / [2⋅2].

+3 = 3

(-3)^{2} = 9

-4⋅2⋅5 = -40

2⋅2 = 4

9 - 40 = -31

√-31 = √31i

Imaginary Number i

So x = [3 ± √31i]/4.

## Formula

### Formula

For a quadratic equation

ax^{2} + bx + c = 0

(a ≠ 0),

x = [-b ± √b^{2} - 4ac] / 2a.

The discriminant, D, is the number

inside the radical sign:

b^{2} - 4ac.

### Nature of the Roots

Recall that

the discriminant D determines

the nature of the roots.

If you know what a complex number is,

you can describe the last case differently.

If D is minus (< 0),

then the quadratic equation has

two complex roots.

Two complex roots are not real roots.

So the previous word,

no real roots,

is true.

## Example 2

### Example

### Solution

The given quadratic equation is

1x^{2} + 3x + 4 = 0.

a = 1

b = +3

c = +4

Then D = 3^{2} - 4⋅1⋅4.

3^{2} = 9

-4⋅1⋅4 = -16

9 - 16 = -7

D = -7

D is minus.

Then the quadratic equation has

two complex roots.

So

two complex roots

is the answer.