ĐK \(x^2+3x\ge0\Rightarrow\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)
\(-x^2-3x+10=3\sqrt{x^2+3x}\Rightarrow x^2+3x+3\sqrt{x^2+3x}-10=0\)
Đặt \(\sqrt{x^2+3x}=t>0\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+3x}=2\Rightarrow x^2+3x-4=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)