Answer
$(-\infty,2)\cup(8,\infty)$
Work Step by Step
$-2|5-x| \lt -6\qquad $ ... divide with $(-2)$
... $\div $(negative) $\Rightarrow $ change direction
$|5-x|\gt 3$
... $|a-b|=|b-a|$
$|x-5|\gt 3$
... $|u| \gt c $ is equivalent to ($ u\lt -c $) or ($ u\gt c $)
$ \begin{array}{lllll}
x-5 \lt-3 & /+5 & ...or... & x-5\gt 3 & /+5\\
x\lt 2 & & & x\gt 8 & \\
x\in(-\infty,2) & & or & x\in(8,\infty) & \\
& & & &
\end{array}$
Solution set: $(-\infty,2)\cup(8,\infty)$
