Find the Roots (Zeros) f(x)=(x+2)^2-16

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

Step 1

Set ${(x + 2)}^{2} - 16$ equal to $0$ .

${(x + 2)}^{2} - 16 = 0$

Step 2

Solve for $x$ .

Add $16$ to both sides of the equation.

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

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

$x + 2 = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

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

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

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

$x + 2 = \pm 4$

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.

$x + 2 = 4$

Move all terms not containing $x$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$x = 4 - 2$

Subtract $2$ from $4$ .

$x = 2$

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

$x + 2 = - 4$

Move all terms not containing $x$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$x = - 4 - 2$

Subtract $2$ from $- 4$ .

$x = - 6$

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

$x = 2, - 6$

Step 3