\(2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\\ 2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\\ \left(3x+1\right)\left(6x+2-x+2\right)=0\\ \left(3x+1\right)\left(5x+4\right)=0\\ \left[{}\begin{matrix}3x+1=0\\5x+4=0\end{matrix}\right.\\ \left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{4}{5}\end{matrix}\right.\)