Learn how to find complex solutions of a quadratic equation with these two examples.

**Example #1:**

Solve 8x^{2} + 200 = 0

Subtract 200 from each side of the equation

8x^{2} + 200 - 200 = 0 - 200

8x^{2} = -200

Divide each side of the equation by 8

(8x^{2})/8 = -200/8

x^{2} = -25

x = ±√(-25)

x = ±√[(25)×-1]

x = ±[√(25) × √(-1)]

x = ±(5 × i)

x = ±5i

x_{1} = 5i

x_{2} = -5i

**Check for 5i**

8(5i)^{2} + 200 = 0

8(25i^{2}) + 200 = 0

8(-25) + 200 = 0

-200 + 200 = 0

**0 = 0**

**Check for -5i**

8(-5i)^{2} + 200 = 0

8(25i^{2}) + 200 = 0

8(-25) + 200 = 0

-200 + 200 = 0

**0 = 0**

**Example #2:**

Solve x^{2} + 4x + 5 = 0

Because this time there is a linear term (4x), you must solve it either using the quadratic formula or by completing the square.

Let us solve by completing the square

Solve x^{2} + 4x + 5 = 0

Isolate the quadratic and the linear term.

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

x^{2} + 4x = -5

Complete the square

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

(x + 2)^{2} = -5 + 4

(x + 2)^{2} = -1

x + 2 = ±i

Subtract 2 from each side of the equation

x + 2 - 2 = ±i - 2

x = ±i - 2

x_{1} = i - 2

x_{2} = -i - 2

**Check for i - 2**

(i - 2)^{2} + 4(i - 2) + 5 = 0

Factor out (i - 2)

(i - 2)[(i - 2) + 4] + 5 = 0

(i - 2)(i + 2) + 5 = 0

i^{2} - 2^{2} + 5 = 0

-1 - 4 + 5 = 0

-5 + 5 = 0

**0 = 0**

**Check for -i - 2**

(-i - 2)^{2} + 4(-i - 2) + 5 = 0

Factor out (-i - 2)

(-i - 2)[(-i - 2) + 4] + 5 = 0

(-i - 2)(-i + 2) + 5 = 0

(-i)^{2} - 2^{2} + 5 = 0

(i)^{2} - 2^{2} + 5 = 0

-1 - 4 + 5 = 0

-5 + 5 = 0

**0 = 0**