Solve for x (x+4)^2-4=16

${(x + 4)}^{2} - 4 = 16$

Add $4$ to both sides of the equation.

${(x + 4)}^{2} = 16 + 4$

Add $16$ and $4$ .

${(x + 4)}^{2} = 20$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x + 4 = \pm \sqrt{20}$

Simplify $\pm \sqrt{20}$ .

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x + 4 = \pm \sqrt{4 (5)}$

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

$x + 4 = \pm \sqrt{2^{2} \cdot 5}$

Pull terms out from under the radical.

$x + 4 = \pm 2 \sqrt{5}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x + 4 = 2 \sqrt{5}$

Subtract $4$ from both sides of the equation.

$x = 2 \sqrt{5} - 4$

Next, use the negative value of the $\pm$ to find the second solution.

$x + 4 = - 2 \sqrt{5}$

Subtract $4$ from both sides of the equation.

$x = - 2 \sqrt{5} - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2 \sqrt{5} - 4, - 2 \sqrt{5} - 4$

The result can be shown in multiple forms.

Exact Form:

$x = 2 \sqrt{5} - 4, - 2 \sqrt{5} - 4$

Decimal Form:

$x = 0.47213595 \dots, - 8.47213595 \dots$