Solve Using the Quadratic Formula 2x^2-2x+7=0

$2 x^{2} - 2 x + 7 = 0$

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 $a = 2$ , $b = - 2$ , and $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (2 \cdot 7)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 2 \cdot 7}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 7$ .

Multiply $- 4$ by $2$ .

$x = \frac{2 \pm \sqrt{4 - 8 \cdot 7}}{2 \cdot 2}$

Multiply $- 8$ by $7$ .

$x = \frac{2 \pm \sqrt{4 - 56}}{2 \cdot 2}$

Subtract $56$ from $4$ .

$x = \frac{2 \pm \sqrt{- 52}}{2 \cdot 2}$

Rewrite $- 52$ as $- 1 (52)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 52}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (52)}$ as $\sqrt{- 1} \cdot \sqrt{52}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{52}}{2 \cdot 2}$

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

$x = \frac{2 \pm i \cdot \sqrt{52}}{2 \cdot 2}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (13)}}{2 \cdot 2}$

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

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 13}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{13})}{2 \cdot 2}$

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{13}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{2 \pm 2 i \sqrt{13}}{4}$

Simplify $\frac{2 \pm 2 i \sqrt{13}}{4}$ .

$x = \frac{1 \pm i \sqrt{13}}{2}$