Ta có: \(x^2-8x+12=0\)
\(\Rightarrow x^2-6x-2x+12=0\)
\(\Rightarrow\left(x^2-6x\right)-\left(2x-12\right)=0\)
\(\Rightarrow x\left(x-6\right)-2\left(x-6\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=6\)