\(6x^2-5x-6=0.\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)\left(x+\dfrac{2}{3}\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=0.\\x+\dfrac{2}{3}=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}.\\x=-\dfrac{2}{3}.\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{3}{2};-\dfrac{2}{3}\right\}.\)