Solve Using the Quadratic Formula n^2+2n-15=0

$n^{2} + 2 n - 15 = 0$

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 $c = - 15$ into the quadratic formula and solve for $n$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 15)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$n = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 15$ .

Multiply $- 4$ by $1$ .

$n = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$n = \frac{- 2 \pm \sqrt{4 + 60}}{2 \cdot 1}$

Add $4$ and $60$ .

$n = \frac{- 2 \pm \sqrt{64}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$n = \frac{- 2 \pm \sqrt{8^{2}}}{2 \cdot 1}$

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

$n = \frac{- 2 \pm 8}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{- 2 \pm 8}{2}$

Simplify $\frac{- 2 \pm 8}{2}$ .

$n = - 1 \pm 4$

Step 4

The final answer is the combination of both solutions.

$n = 3, - 5$