Answer
The solution set is $(-1,4)$
Work Step by Step
$3x^2-3x<2x^2+4\hspace{0.7cm}{\color{blue}{\text{Given equation}}}$
$\Rightarrow 3x^2-3x-2x^2-4<0$
$\Rightarrow x^2-3x-4<0$
$\Rightarrow (x-4)(x+1)<0\hspace{0.7cm}{\color{blue}{\text{Factor}}}$
The factors of the left-hand side are $x-4$ and $x+1$.
These factors are zero when $x=4$ and $x=-1$.
These numbers divide the real line into the intervals
$(-\infty,-1),(-1,4),(4,\infty)$
