Answer
The inequality is valid for values less than -2/3 and more than 1.5 (not including them) i.e. $(-\infty,-\frac{2}{3})\cap (1.5,\infty)$
Work Step by Step
First, we are going to set the right side to zero and factor to find the x-intercepts:
$6(x^2-1)=5x$
$6x^2-6-5x=5x-5x$
$6x^2-5x-6=0$
$(3x+2)(2x-3)=0$
$x_1=-2/3$
$x_2=1.5$
These are the critical points. We are going to take three values: one less than -2/3, one between -2/3 and 1.5, and one more than 1.5 to test in the original equation and check if the inequality is true or not:
First test with a value less than -2/3:
$6((-1)^2-1)>5(-1)$
$6(1-1)>-1$
$6(0)>-1$
$0>-1 \rightarrow \text{ TRUE}$
Second test with a value between -2/3 and 1.5:
$6(0^2-1)>5(0)$
$6(0-1)>0$
$6(-1)>0$
$-6>0 \rightarrow \text{ FALSE}$
Third test with a value more than 1.5:
$6(2^2-1)>5(2)$
$6(4-1)>10$
$6(3)>10$
$18>0 \rightarrow \text{ TRUE}$
These tests show that the inequality $6(x^2-1)>5x$ is valid for values less than -2/3 and more than 1.5 (not including them) i.e. $(-\infty,-\frac{2}{3})\cap (1.5,\infty)$