$x.(x+7)=0⇔\left[ \begin{array}{l}x=0\\x+7=0\Rightarrow x=-7\end{array} \right.$
$(x+12)(3-x)=0⇔\left[ \begin{array}{l}x+12=0\Rightarrow x=-12\\3-x=0\Rightarrow x=3\end{array} \right.$
$(-x+5)(3-x)=0⇔\left[ \begin{array}{l}-x+5=0\Rightarrow x=5\\3-x=0\Rightarrow x=3\end{array} \right.$
$x(2+x)(7-x)=0⇔\left[ \begin{array}{l}x=0\\2+x=0\Rightarrow x=-2\\7-x=0\Rightarrow x=7\end{array} \right.$
$(x-1)(x+2)(-x-3)=0⇔\left[ \begin{array}{l}x-1=0 \Rightarrow x=1\\x+2=0\Rightarrow x=-2\\-x-3=0\Rightarrow x=-3\end{array} \right.$
