Solve Using the Quadratic Formula y^2-2y+5=0

$y^{2} - 2 y + 5 = 0$

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 $a = 1$ , $b = - 2$ , and $c = 5$ into the quadratic formula and solve for $y$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 5$ .

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 5}}{2 \cdot 1}$

Multiply $- 4$ by $5$ .

$y = \frac{2 \pm \sqrt{4 - 20}}{2 \cdot 1}$

Subtract $20$ from $4$ .

$y = \frac{2 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$y = \frac{2 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$y = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

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

$y = \frac{2 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$y = \frac{2 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

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

$y = \frac{2 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$y = \frac{2 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{2 \pm 4 i}{2}$

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

$y = 1 \pm 2 i$