\(\sqrt{x+2}=3+2x\Leftrightarrow\left(\sqrt{x+2}\right)^2=\left(3+2x\right)^2\Leftrightarrow x+2=9+12x+4x^2\Leftrightarrow x-12x+2-9-4x^2=0\Leftrightarrow-11x-4x^2-7=0\Leftrightarrow-4x^2-4x-7x-7=0\Leftrightarrow-4x\left(x+1\right)-7\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(4x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\4x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{-7}{4}=-1,75\end{matrix}\right.\left(TMĐK\right)\)ĐKXĐ: x+2\(\ge\)0\(\Leftrightarrow x\ge-2\). Vậy tập nghiệm của pt:\(S=\left\{-1;-1,75\right\}\)
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