Solve Using the Quadratic Formula x^2+17=2x

$x^{2} + 17 = 2 x$

Step 1

Subtract $2 x$ from both sides of the equation.

$x^{2} + 17 - 2 x = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = - 2$ , and $c = 17$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 17)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 17}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 17$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 17}}{2 \cdot 1}$

Multiply $- 4$ by $17$ .

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

Subtract $68$ from $4$ .

$x = \frac{2 \pm \sqrt{- 64}}{2 \cdot 1}$

Rewrite $- 64$ as $- 1 (64)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 1}$

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

$x = \frac{2 \pm i \cdot \sqrt{64}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 1}$

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

$x = \frac{2 \pm i \cdot 8}{2 \cdot 1}$

Move $8$ to the left of $i$ .

$x = \frac{2 \pm 8 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 8 i}{2}$

Simplify $\frac{2 \pm 8 i}{2}$ .

$x = 1 \pm 4 i$