$\frac{-6}{\sqrt{x}+3}<\frac{-1}{2}<=>\frac{1}{\sqrt{x}+3}>\frac{1}{12}<=>\sqrt{x}+3<12<=>\sqrt{x}<9<=>0\leqslant x<81, x\neq 9$
$\frac{-6}{\sqrt{x}+3}<\frac{-1}{2}<=>\frac{1}{\sqrt{x}+3}>\frac{1}{12}<=>\sqrt{x}+3<12<=>\sqrt{x}<8<=>0\leqslant x<64, x\neq 9$
$\frac{-6}{\sqrt{x}+3}<\frac{-1}{2}<=>\frac{1}{\sqrt{x}+3}>\frac{1}{12}<=>\sqrt{x}+3<12<=>\sqrt{x}<
9<=>0\leqslant x<
81, x\neq 9$