Solve Using the Quadratic Formula 2x^2-6x=-8

$2 x^{2} - 6 x = - 8$

Step 1

Add $8$ to both sides of the equation.

$2 x^{2} - 6 x + 8 = 0$

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 $a = 2$ , $b = - 6$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (2 \cdot 8)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot 8}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 8$ .

Multiply $- 4$ by $2$ .

$x = \frac{6 \pm \sqrt{36 - 8 \cdot 8}}{2 \cdot 2}$

Multiply $- 8$ by $8$ .

$x = \frac{6 \pm \sqrt{36 - 64}}{2 \cdot 2}$

Subtract $64$ from $36$ .

$x = \frac{6 \pm \sqrt{- 28}}{2 \cdot 2}$

Rewrite $- 28$ as $- 1 (28)$ .

$x = \frac{6 \pm \sqrt{- 1 \cdot 28}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (28)}$ as $\sqrt{- 1} \cdot \sqrt{28}$ .

$x = \frac{6 \pm \sqrt{- 1} \cdot \sqrt{28}}{2 \cdot 2}$

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

$x = \frac{6 \pm i \cdot \sqrt{28}}{2 \cdot 2}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{6 \pm i \cdot \sqrt{4 (7)}}{2 \cdot 2}$

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

$x = \frac{6 \pm i \cdot \sqrt{2^{2} \cdot 7}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{6 \pm i \cdot (2 \sqrt{7})}{2 \cdot 2}$

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

$x = \frac{6 \pm 2 i \sqrt{7}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{6 \pm 2 i \sqrt{7}}{4}$

Simplify $\frac{6 \pm 2 i \sqrt{7}}{4}$ .

$x = \frac{3 \pm i \sqrt{7}}{2}$