Solve Using the Quadratic Formula x^2+14x+13=0

$x^{2} + 14 x + 13 = 0$

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 $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (1 \cdot 13)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot 1 \cdot 13}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 13$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot 13}}{2 \cdot 1}$

Multiply $- 4$ by $13$ .

$x = \frac{- 14 \pm \sqrt{196 - 52}}{2 \cdot 1}$

Subtract $52$ from $196$ .

$x = \frac{- 14 \pm \sqrt{144}}{2 \cdot 1}$

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

$x = \frac{- 14 \pm \sqrt{12^{2}}}{2 \cdot 1}$

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

$x = \frac{- 14 \pm 12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 14 \pm 12}{2}$

Simplify $\frac{- 14 \pm 12}{2}$ .

$x = - 7 \pm 6$

Step 4

The final answer is the combination of both solutions.

$x = - 1, - 13$