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
ax2 + bx + c = 0
(a ≠ 0),

x = [-b ± √b2 - 4ac] / 2a.

Let's use this
to find the complex roots.

## Example 1

### Solution

2x2 - 3x + 5 = 0.

a = 2
b = -3
c = +5

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

ax2 + bx + c = 0
(a ≠ 0),

x = [-b ± √b2 - 4ac] / 2a.

The discriminant, D, is the number
b2 - 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),
two complex roots.

Two complex roots are not real roots.
So the previous word,
no real roots,
is true.

## Example 2

### Solution

1x2 + 3x + 4 = 0.

a = 1
b = +3
c = +4

Then D = 32 - 4⋅1⋅4.

32 = 9
-4⋅1⋅4 = -16

9 - 16 = -7

D = -7

D is minus.

two complex roots.

So
two complex roots