Solve Using the Quadratic Formula x^2+4x+4=12

$x^{2} + 4 x + 4 = 12$

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $12$ from both sides of the equation.

$x^{2} + 4 x + 4 - 12 = 0$

Subtract $12$ from $4$ .

$x^{2} + 4 x - 8 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = 4$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot - 8)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 8$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$x = \frac{- 4 \pm \sqrt{16 + 32}}{2 \cdot 1}$

Add $16$ and $32$ .

$x = \frac{- 4 \pm \sqrt{48}}{2 \cdot 1}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$x = \frac{- 4 \pm \sqrt{16 (3)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 4 \pm \sqrt{4^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm 4 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 4 \sqrt{3}}{2}$

Simplify $\frac{- 4 \pm 4 \sqrt{3}}{2}$ .

$x = - 2 \pm 2 \sqrt{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - 2 \pm 2 \sqrt{3}$

Decimal Form:

$x = 1.46410161 \dots, - 5.46410161 \dots$