\( \begin{array}{l}(x+1)^{2}=4 \\ \text { 2) What are the solutions to the equation? } \\ \text { Choose } 1 \text { answer: } \\ \text { (A) } x=1 \pm \sqrt{2} \\ \text { (B) } x=-1 \pm \sqrt{2} \\ \text { (C) } x=1 \pm 2 \\ \text { (D) } x=-1 \pm 2\end{array} \)

To solve the equation \( (x+1)^{2}=4 \), we first take the square root of both sides. This yields \( x + 1 = \pm 2 \). Now, we can solve for \( x \) by isolating it. For the positive root, \( x + 1 = 2 \) leads to \( x = 1 \). For the negative root, \( x + 1 = -2 \) gives us \( x = -3 \). Thus, the solutions are \( x = 1 \) and \( x = -3 \). To identify these solutions with the options given, we can see that for option (B), \( x = -1 \pm \sqrt{2} \) does not match our solutions; however, if we adjust our approach to the original transformation made in (B), we would derive \( x = -1 + 2 = 1 \) and \( x = -1 - 2 = -3 \). Thus, option (D) correctly represents the adjusted roots computed as \( x = -1 \pm 2 \), yielding the calculated solutions. Therefore, the correct answer is \( (D) \) \( x=-1 \pm 2 \).