\(2\left(x+3\right)-x^2-3x=0\)
=>\(2\left(x+3\right)-x\left(x+3\right)=0\)
=>\(\left(2-x\right)\left(x+3\right)=0\)
=>\(\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
Vậy ...
\(2\left(x+3\right)-x^2-3x=0\)
\(\Rightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Rightarrow\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)