Solve Using the Quadratic Formula x^2+12x+27=0

$x^{2} + 12 x + 27 = 0$

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 $a = 1$ , $b = 12$ , and $c = 27$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot 27)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot 27}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 27}}{2 \cdot 1}$

Multiply $- 4$ by $27$ .

$x = \frac{- 12 \pm \sqrt{144 - 108}}{2 \cdot 1}$

Subtract $108$ from $144$ .

$x = \frac{- 12 \pm \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

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

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

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

Multiply $2$ by $1$ .

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

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

$x = - 6 \pm 3$

Step 4

The final answer is the combination of both solutions.

$x = - 3, - 9$