Elementary Algebra

$a=3$
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.