$4x^2 - 12x + 9 = 9$
$\Rightarrow 4x^2 - 12x = 0$
$\Rightarrow x(4x - 12) = 0$
\(\left[ \begin{array}{l}x=0\\4x-12 = 0 \end{array} \right.\)
\(\left[ \begin{array}{l}x=0\\4x = 12\end{array} \right.\)
\(\left[ \begin{array}{l}x=0\\x = 3 \end{array} \right.\)
Vậy `x ∈ {0 ; 3}`