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