Solve Using the Quadratic Formula g^2+4g-32=0

$g^{2} + 4 g - 32 = 0$

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 $c = - 32$ into the quadratic formula and solve for $g$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 32)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$g = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 32}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 32$ .

Multiply $- 4$ by $1$ .

$g = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 32}}{2 \cdot 1}$

Multiply $- 4$ by $- 32$ .

$g = \frac{- 4 \pm \sqrt{16 + 128}}{2 \cdot 1}$

Add $16$ and $128$ .

$g = \frac{- 4 \pm \sqrt{144}}{2 \cdot 1}$

Rewrite $144$ as $12^{2}$ .

$g = \frac{- 4 \pm \sqrt{12^{2}}}{2 \cdot 1}$

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

$g = \frac{- 4 \pm 12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$g = \frac{- 4 \pm 12}{2}$

Simplify $\frac{- 4 \pm 12}{2}$ .

$g = - 2 \pm 6$

Step 4

The final answer is the combination of both solutions.

$g = 4, - 8$