Solve for x x+3 = square root of 16

$x + 3 = \sqrt{16}$

Simplify $\sqrt{16}$ .

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

$x + 3 = \sqrt{4^{2}}$

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

$x + 3 = 4$

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

Subtract $3$ from both sides of the equation.

$x = 4 - 3$

Subtract $3$ from $4$ .

$x = 1$