This problem has two answer to it, Help!

as you should already know, an absolute value expression, is in effect a piece-wise expression, since it has a ± versions.
[tex] \bf \cfrac{|-2-2r|}{3}=4\implies
\begin{cases}
\cfrac{+(-2-2r)}{3}=4\\\\
\cfrac{-(-2-2r)}{3}=4
\end{cases}\\\\[-0.35em]
~\dotfill\\\\
\cfrac{-2-2r}{3}=4\implies -2-2r=12\implies -2-12=2r\implies -14=2r
\\\\\\
\cfrac{-14}{2}=r\implies \boxed{-7=r}\\\\[-0.35em]
~\dotfill\\\\
\cfrac{-(-2-2r)}{3}=4\implies \cfrac{-2-2r}{3}=-4\implies -2-2r=-12
\\\\\\
-2+12=2r\implies 10=2r\implies \cfrac{10}{2}=r\implies \boxed{5=r} [/tex]