Solve for w 2w^2-4w+5=0

$2 w^{2} - 4 w + 5 = 0$

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 $a = 2$ , $b = - 4$ , and $c = 5$ into the quadratic formula and solve for $w$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (2 \cdot 5)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

$w = \frac{4 \pm \sqrt{16 - 4 \cdot 2 \cdot 5}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 5$ .

Multiply $- 4$ by $2$ .

$w = \frac{4 \pm \sqrt{16 - 8 \cdot 5}}{2 \cdot 2}$

Multiply $- 8$ by $5$ .

$w = \frac{4 \pm \sqrt{16 - 40}}{2 \cdot 2}$

Subtract $40$ from $16$ .

$w = \frac{4 \pm \sqrt{- 24}}{2 \cdot 2}$

Rewrite $- 24$ as $- 1 (24)$ .

$w = \frac{4 \pm \sqrt{- 1 \cdot 24}}{2 \cdot 2}$

Rewrite $\sqrt{- 1 (24)}$ as $\sqrt{- 1} \cdot \sqrt{24}$ .

$w = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{24}}{2 \cdot 2}$

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

$w = \frac{4 \pm i \cdot \sqrt{24}}{2 \cdot 2}$

Rewrite $24$ as $2^{2} \cdot 6$ .

Factor $4$ out of $24$ .

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

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

$w = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 6}}{2 \cdot 2}$

Pull terms out from under the radical.

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

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

$w = \frac{4 \pm 2 i \sqrt{6}}{2 \cdot 2}$

Multiply $2$ by $2$ .

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

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

$w = \frac{2 \pm i \sqrt{6}}{2}$

Step 4

The final answer is the combination of both solutions.

$w = \frac{2 + i \sqrt{6}}{2}, \frac{2 - i \sqrt{6}}{2}$