`x²` x `(x - 3) - 12 + 4x = 0`
`⇔ x²` x `(x - 3) + 4x - 12 = 0`
`⇔ x²` x `(x - 3) + 4 (x - 3) = 0`
`⇔ (x² + 4) (x - 3) = 0`
`⇔` \(\left[ \begin{array}{l}x^2 + 4=0\\x - 3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x^2 = -4 (vô lí vì x^2 ≥ 0) \\x = 3\end{array} \right.\)
Vậy `x = 3`