Answer
$(-\infty,-2)\cup(6,\infty)$
Work Step by Step
Rewrite as
$|2-x| \gt 4$
... also, $|a-b|=|b-a|$
$|x-2| \gt 4$
... $|u| \gt c $ is equivalent to ($ u\lt -c $) or ($ u\gt c $)
$ \begin{array}{lllll}
x-2\lt-4 & /+2 & ...or... & x-2\gt 4 & /+2\\
x\lt-2 & & & x \gt 6 & \\
x\in(-\infty,-2) & & or & x\in(6,\infty) & \\
& & & &
\end{array} $
Solution set: $(-\infty,-2)\cup(6,\infty)$
