Find All Complex Solutions x^2-4x+68=0

$x^{2} - 4 x + 68 = 0$

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 $a = 1$ , $b = - 4$ , and $c = 68$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $68$ .

$x = \frac{4 \pm \sqrt{16 - 272}}{2 \cdot 1}$

Subtract $272$ from $16$ .

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

Rewrite $- 256$ as $- 1 (256)$ .

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

Rewrite $\sqrt{- 1 (256)}$ as $\sqrt{- 1} \cdot \sqrt{256}$ .

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

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

$x = \frac{4 \pm i \cdot \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

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

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

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

Move $16$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

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

$x = 2 \pm 8 i$