Solve Using the Quadratic Formula x^2-6x-27=0

$x^{2} - 6 x - 27 = 0$

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 $a = 1$ , $b = - 6$ , and $c = - 27$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 27)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 27}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 27$ .

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 27}}{2 \cdot 1}$

Multiply $- 4$ by $- 27$ .

$x = \frac{6 \pm \sqrt{36 + 108}}{2 \cdot 1}$

Add $36$ and $108$ .

$x = \frac{6 \pm \sqrt{144}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{12^{2}}}{2 \cdot 1}$

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

$x = \frac{6 \pm 12}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 12}{2}$

Simplify $\frac{6 \pm 12}{2}$ .

$x = 3 \pm 6$

Step 4

The final answer is the combination of both solutions.

$x = 9, - 3$