Find All Complex Solutions x^2-16x+68=0

$x^{2} - 16 x + 68 = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = 1$ , $b = - 16$ , and $c = 68$ into the quadratic formula and solve for $x$ .

$\frac{16 \pm \sqrt{{(- 16)}^{2} - 4 \cdot (1 \cdot 68)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 16$ to the power of $2$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 1 \cdot 68}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 68$ .

Multiply $- 4$ by $1$ .

$x = \frac{16 \pm \sqrt{256 - 4 \cdot 68}}{2 \cdot 1}$

Multiply $- 4$ by $68$ .

$x = \frac{16 \pm \sqrt{256 - 272}}{2 \cdot 1}$

Subtract $272$ from $256$ .

$x = \frac{16 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$x = \frac{16 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \frac{16 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{16 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{16 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{16 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$x = \frac{16 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{16 \pm 4 i}{2}$

Simplify $\frac{16 \pm 4 i}{2}$ .

$x = 8 \pm 2 i$