Solve Using the Quadratic Formula x^2-12x-45=0

$x^{2} - 12 x - 45 = 0$

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 $c = - 45$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (1 \cdot - 45)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 1 \cdot - 45}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 45$ .

Multiply $- 4$ by $1$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot - 45}}{2 \cdot 1}$

Multiply $- 4$ by $- 45$ .

$x = \frac{12 \pm \sqrt{144 + 180}}{2 \cdot 1}$

Add $144$ and $180$ .

$x = \frac{12 \pm \sqrt{324}}{2 \cdot 1}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{12 \pm \sqrt{18^{2}}}{2 \cdot 1}$

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

$x = \frac{12 \pm 18}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{12 \pm 18}{2}$

Simplify $\frac{12 \pm 18}{2}$ .

$x = 6 \pm 9$

Step 4

The final answer is the combination of both solutions.

$x = 15, - 3$