`(2x-3)^2 - 4x(x+1)^2 =12`
`<=>(2x)^2 - 2.2x.3 + 3^2 - 4x(x^2 + 2x+1)=12`
`<=>4x^2 - 12x+9 - 4x^3 - 8x^2 - 4x=12`
`<=>(4x^2 - 8x^2) + (-12x-4x) = 12-9`
`<=>-4x^2 - 16x = 3`
`<=>-4x^2 - 16x-3=0`
`<=>` \(\left[{}\begin{matrix}x=\dfrac{-4+\sqrt{13}}{2}\\x=\dfrac{-4-\sqrt{13}}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{-4+\sqrt{13}}{2}\) hoặc \(x=\dfrac{-4-\sqrt{13}}{2}\)