Answer
$w=2$
Work Step by Step
In order to get rid of the square root, square both sides:
\begin{align*}
(2\sqrt{w-1})^2&=(\sqrt{w+2})^2\\
2^2(w-1)&=w+2\\
4(w-1)&=w+2\\
4w-4&=w+2
\end{align*}
Move the $w$ terms to one side:
$4w-w=2+4$
$3w=6$
$w=6/3$
$w=2$
Test the solution to make sure that it is not extraneous:
$2\sqrt{2-1}=\sqrt{2+2}$
$2\sqrt{1}=\sqrt{4}$
$2\cdot 1=2$
$2=2$
The solution works.