#### Answer

The solution set is $\left\{-2\right\}$.

#### Work Step by Step

Square both sides to obtain:
$x^2=(2\sqrt{-x-1})^2$
Use the rule $(ab)^m = a^mb^m$ to obtain:
$x^2=2^2(\sqrt{-x-1})^2
\\x^2=4(-x-1)
\\x^2=-4x-4$
Add $4x$ and $4$ on both sides of the equation to obtain:
$x^2+4x+4 = -4x-4+4x+4
\\x^2+4x+4=0$
Factor the trinomial to obtain:
$(x+2)(x+2) = 0$
Equate each factor to zero, and then solve each equation to obtain:
$\begin{array}{ccc}
&x+2=0 &\text{ or } &x+2=0
\\&x=-2 &\text{ or } &x=-2
\end{array}$
Thus, the solution set is $\left\{-2\right\}$.