Solve Using the Quadratic Formula n^2+5n-14=0

$n^{2} + 5 n - 14 = 0$

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 $a = 1$ , $b = 5$ , and $c = - 14$ into the quadratic formula and solve for $n$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 14)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$n = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 14}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 14$ .

Multiply $- 4$ by $1$ .

$n = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 14}}{2 \cdot 1}$

Multiply $- 4$ by $- 14$ .

$n = \frac{- 5 \pm \sqrt{25 + 56}}{2 \cdot 1}$

Add $25$ and $56$ .

$n = \frac{- 5 \pm \sqrt{81}}{2 \cdot 1}$

Rewrite $81$ as $9^{2}$ .

$n = \frac{- 5 \pm \sqrt{9^{2}}}{2 \cdot 1}$

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

$n = \frac{- 5 \pm 9}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{- 5 \pm 9}{2}$

Step 4

The final answer is the combination of both solutions.

$n = 2, - 7$