Answer
The solution set is $(-\infty,-3)\cup(6,\infty)$
Work Step by Step
$x^2>3(x+6)\hspace{0.7cm}{\color{blue}{\text{Given equation}}}$
$\Rightarrow x^2-3(x+6)>0$
$\Rightarrow x^2-3x-18>0$
$\Rightarrow (x+3)(x-6)\ge0\hspace{0.7cm}{\color{blue}{\text{Factor}}}$
The factors of the left-hand side are $x+3$ and $x-6$.
These factors are zero when $x=-3$ and $x=6$.
These numbers divide the real line into the intervals
$(-\infty,-3),(-3,6),(6,\infty)$
From the diagram and hence the inequality involves $>$, the endpoints of the intervals do not satisfy the inequality.
The solution set is $(-\infty,-3)\cup(6,\infty)$
