$(2x-12).(x²-81)=0$
$⇔2.(x-6).(x²-81)=0$
$⇔\left[ \begin{array}{l}x-6=0\\x²-81=0\end{array} \right.⇔\left[ \begin{array}{l}x=0+6\\x²=81\end{array} \right.$
$⇔\left[ \begin{array}{l}x=6\\x²=(±9)²\end{array} \right.⇔\left[ \begin{array}{l}x=6\\x=±9\end{array} \right.$
Vậy $x∈${$6;±9$}.