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

$y^{2} + 2 y - 8 = 0$

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 $c = - 8$ into the quadratic formula and solve for $y$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 8)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$y = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 8$ .

Multiply $- 4$ by $1$ .

$y = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$y = \frac{- 2 \pm \sqrt{4 + 32}}{2 \cdot 1}$

Add $4$ and $32$ .

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

$y = - 1 \pm 3$

Step 4

The final answer is the combination of both solutions.

$y = 2, - 4$