Solve Using the Quadratic Formula x^2-14x-15=0

$x^{2} - 14 x - 15 = 0$

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 $a = 1$ , $b = - 14$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{14 \pm \sqrt{{(- 14)}^{2} - 4 \cdot (1 \cdot - 15)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 14$ to the power of $2$ .

$x = \frac{14 \pm \sqrt{196 - 4 \cdot 1 \cdot - 15}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{14 \pm \sqrt{196 - 4 \cdot - 15}}{2 \cdot 1}$

Multiply $- 4$ by $- 15$ .

$x = \frac{14 \pm \sqrt{196 + 60}}{2 \cdot 1}$

Add $196$ and $60$ .

$x = \frac{14 \pm \sqrt{256}}{2 \cdot 1}$

Rewrite $256$ as $16^{2}$ .

$x = \frac{14 \pm \sqrt{16^{2}}}{2 \cdot 1}$

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

$x = \frac{14 \pm 16}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{14 \pm 16}{2}$

Simplify $\frac{14 \pm 16}{2}$ .

$x = 7 \pm 8$

Step 4

The final answer is the combination of both solutions.

$x = 15, - 1$