Answer
$a=3$
Work Step by Step
Squaring both sides of the inequality and using the properties of radicals, the solution to the given equation is
\begin{array}{l}\require{cancel}\left(
\sqrt{-3a+10}
\right)^2=\left(
a-2
\right)^2
\\\\
-3a+10=(a)^2+2(a)(-2)+(-2)^2
\\\\
-3a+10=a^2-4a+4
\\\\
0=a^2+(-4a+3a)+(4-10)
\\\\
a^2-a-6=0
\\\\
(a-3)(a+2)=0
\\\\
a=\{ -2,3 \}
.\end{array}
Upon checking, only $
a=3
$ satisfies the original equation.
