Solve Using the Quadratic Formula n^2+2n-24=0

$n^{2} + 2 n - 24 = 0$

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 $c = - 24$ into the quadratic formula and solve for $n$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 24)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$n = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 24$ .

Multiply $- 4$ by $1$ .

$n = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 24}}{2 \cdot 1}$

Multiply $- 4$ by $- 24$ .

$n = \frac{- 2 \pm \sqrt{4 + 96}}{2 \cdot 1}$

Add $4$ and $96$ .

$n = \frac{- 2 \pm \sqrt{100}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$n = \frac{- 2 \pm \sqrt{10^{2}}}{2 \cdot 1}$

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

$n = \frac{- 2 \pm 10}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{- 2 \pm 10}{2}$

Simplify $\frac{- 2 \pm 10}{2}$ .

$n = - 1 \pm 5$

Step 4

The final answer is the combination of both solutions.

$n = 4, - 6$