\(3x^2-5x-2=0\\ 3x^2+x-6x-2=0\\ x\left(3x+1\right)-2\left(3x+1\right)=0\\ \left(3x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=-1\\x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{-1}{3};2\right\}\)
\(\Delta=25-4.\left(-2\right).3=49\Rightarrow\left\{{}\begin{matrix}x_1=\frac{5+7}{6}=2\\x_2=\frac{5-7}{6}=\frac{-1}{3}\end{matrix}\right.\)