Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
=>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>\(\left[\begin{array}{l}x-2=0\\ 3x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=\frac13\end{array}\right.\)