\(\Leftrightarrow3x+9-x^2-3x=0\\ \Leftrightarrow\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
3(x+3)-x^2-3x=0
3(x+3)-x(x+3)=0
(3-x)(x+3)=0
TH1: 3-x=0 TH2: x+3=0
x=3 x=-3
Vậy x\(\in\){-3;3}