\(x\ge-\frac{10}{3}\)
\(x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0\)
\(\Leftrightarrow\left(x+3\right)^2-\left(\sqrt{3x+10}-1\right)^2=0\)
\(\Leftrightarrow\left(x+4-\sqrt{3x+10}\right)\left(x+2+\sqrt{3x+10}\right)=0\)
TH1: \(x+4-\sqrt{3x+10}=0\Leftrightarrow\left\{{}\begin{matrix}x+4\ge0\\\left(x+4\right)^2=3x+10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x^2+5x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
TH2: \(x+2+\sqrt{3x+10}=0\Leftrightarrow\left\{{}\begin{matrix}-x-2\ge0\\\left(-x-2\right)^2=3x+10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=-3\end{matrix}\right.\)
