Answer
$x=-1+\sqrt{2}$ or $x=-1-\sqrt{2}$
Work Step by Step
$(x+1)^{2}=2\qquad$...use the general principle of square roots:
For any real number $k$ and any algebraic expression $X$ :
$\text{If }X^{2}=k,\text{ then }X=\sqrt{k}\text{ or }X=-\sqrt{k}$
$ x+1=\pm\sqrt{2}\qquad$...add $-1$ to both sides.
$ x=-1\pm\sqrt{2}\qquad$... the symbol $\pm$ indicates two solutions.
$x=-1+\sqrt{2}$ or $x=-1-\sqrt{2}$
