Solve Using the Quadratic Formula 2x^2-8x+9=0

$2 x^{2} - 8 x + 9 = 0$

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 $a = 2$ , $b = - 8$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (2 \cdot 9)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 2 \cdot 9}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 9$ .

Multiply $- 4$ by $2$ .

$x = \frac{8 \pm \sqrt{64 - 8 \cdot 9}}{2 \cdot 2}$

Multiply $- 8$ by $9$ .

$x = \frac{8 \pm \sqrt{64 - 72}}{2 \cdot 2}$

Subtract $72$ from $64$ .

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

Rewrite $- 8$ as $- 1 (8)$ .

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

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

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

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

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

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{8 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 2}$

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

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

Pull terms out from under the radical.

$x = \frac{8 \pm i \cdot (2 \sqrt{2})}{2 \cdot 2}$

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

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

Multiply $2$ by $2$ .

$x = \frac{8 \pm 2 i \sqrt{2}}{4}$

Simplify $\frac{8 \pm 2 i \sqrt{2}}{4}$ .

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