Solve Using the Quadratic Formula y^2+4y+13=0

$y^{2} + 4 y + 13 = 0$

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 $a = 1$ , $b = 4$ , and $c = 13$ into the quadratic formula and solve for $y$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 13$ .

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 13}}{2 \cdot 1}$

Multiply $- 4$ by $13$ .

$y = \frac{- 4 \pm \sqrt{16 - 52}}{2 \cdot 1}$

Subtract $52$ from $16$ .

$y = \frac{- 4 \pm \sqrt{- 36}}{2 \cdot 1}$

Rewrite $- 36$ as $- 1 (36)$ .

$y = \frac{- 4 \pm \sqrt{- 1 \cdot 36}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

$y = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{36}}{2 \cdot 1}$

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

$y = \frac{- 4 \pm i \cdot \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$y = \frac{- 4 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 1}$

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

$y = \frac{- 4 \pm i \cdot 6}{2 \cdot 1}$

Move $6$ to the left of $i$ .

$y = \frac{- 4 \pm 6 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm 6 i}{2}$

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

$y = - 2 \pm 3 i$