Solve Using the Quadratic Formula x^2+12x+c

$x^{2} + 12 x + c$

Set $x^{2} + 12 x + c$ equal to $0$ .

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

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

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Factor $4$ out of $144 - 4 c$ .

Factor $4$ out of $144$ .

$x = \frac{- 12 \pm \sqrt{4 (36) - 4 c}}{2 \cdot 1}$

Factor $4$ out of $- 4 c$ .

$x = \frac{- 12 \pm \sqrt{4 (36) + 4 (- c)}}{2 \cdot 1}$

Factor $4$ out of $4 (36) + 4 (- c)$ .

$x = \frac{- 12 \pm \sqrt{4 (36 - c)}}{2 \cdot 1}$

Rewrite $4 (36 - c)$ as $2^{2} (6^{2} - c)$ .

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 12 \pm \sqrt{2^{2} (36 - c)}}{2 \cdot 1}$

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 12 \pm 2 \sqrt{36 - c}}{2}$ .

$x = - 6 \pm \sqrt{36 - c}$