OpenSAT

To solve for $x$, we take the square root of both sides of the equation: $\sqrt{(x+1)^2} = \sqrt{16}$. This gives us $x + 1 = \pm 4$. Solving for $x$ in each case, we get $x = 4 - 1 = 3$ or $x = -4 - 1 = -5$. Of these two solutions, only 3 is given as a choice.