Solve Using the Quadratic Formula x^2+12x-13=0

$x^{2} + 12 x - 13 = 0$

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 $a = 1$ , $b = 12$ , and $c = - 13$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot - 13)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot - 13}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot - 13}}{2 \cdot 1}$

Multiply $- 4$ by $- 13$ .

$x = \frac{- 12 \pm \sqrt{144 + 52}}{2 \cdot 1}$

Add $144$ and $52$ .

$x = \frac{- 12 \pm \sqrt{196}}{2 \cdot 1}$

Rewrite $196$ as $14^{2}$ .

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

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

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

Multiply $2$ by $1$ .

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

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

$x = - 6 \pm 7$

Step 4

The final answer is the combination of both solutions.

$x = 1, - 13$