Solve (x+1)^2-48=0 where x is a real number. Simplify your answer as much as possible. If there is more than one solution , separate them with commas. If there is no solution, click "No solution." x= square

To solve the equation $(x+1)^{2}-48=0$, we can start by moving the constant term to the other side of the equation:<br /><br />$(x+1)^{2} = 48$<br /><br />Next, we take the square root of both sides of the equation:<br /><br />$x+1 = \pm \sqrt{48}$<br /><br />Simplifying further, we have:<br /><br />$x+1 = \pm 4\sqrt{3}$<br /><br />Finally, we solve for x by subtracting 1 from both sides:<br /><br />$x = -1 \pm 4\sqrt{3}$<br /><br />Therefore, the solutions to the equation are $x = -1 + 4\sqrt{3}$ and $x = -1 - 4\sqrt{3}$.