\(PT\Leftrightarrow x^2-x-3\sqrt{x^2-x-2}=0\\ \Leftrightarrow\left(x^2-x-6\right)-3\left(\sqrt{x^2-x-2}-2\right)=0\\ \Leftrightarrow\left(x^2-x-6\right)-\dfrac{3\left(x^2-x-6\right)}{\sqrt{x^2-x-2}+2}=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)\left(1-\dfrac{3}{\sqrt{x^2-x-2}+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\1-\dfrac{3}{\sqrt{x^2-x-2}+2}=0\left(1\right)\end{matrix}\right.\)
Ta có \(\dfrac{3}{\sqrt{x^2-x-2}+2}\le\dfrac{3}{2}\Leftrightarrow1-\dfrac{3}{\sqrt{x^2-x-2}+2}\le-\dfrac{1}{2}< 0\) nên \(\left(1\right)\) vô nghiệm
Vậy pt có nghiệm \(S=\left\{-2;3\right\}\)