Answer
$x$ is in the interval $(-\infty,-\frac{2}{3}) \cup (\frac{3}{2}, \infty)$.
Work Step by Step
$6(x^2-1)>5x$, I subtract $5x$ from both sides, hence $6x^2-5x-6>0$.
The coefficient of $x^2$ is $6$, hence the coefficient of $x$ in the multiplicants must be $6$ and $1$, $2$ and $3$, $-6$ and $-1$ or $-2$ and $-3$. The coefficient of $x^0$ is 6, hence the coefficient of $x^0$ in the multiplicants must be $6$ and $1$, $2$ and $3$, $-6$ and $-1$ or $-2$ and $-3$.
The following combination is good $(3x+2)(2x-3)>0$, hence the solution set is: $x$ is in the interval $(-\infty,-\frac{2}{3}) \cup (\frac{3}{2}, \infty)$. (also see the graph of $(3x+2)(2x-3)$
