Answer
The inequality is valid for values between -2/3 and 1.5 (not including them) i.e. $(-\frac{2}{3},1.5)$
Work Step by Step
First, we are going to set the right side to zero and factor to find the x-intercepts:
$6x^2-5x-6=0$
$(3x+2)(2x-3)=0$
$x_1=-2/3$
$x_2=1.5$
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:
$6(-1)^2<6+5(-1)$
$6(1)<6-5$
$6<1 \rightarrow \text{ FALSE}$
Second test with a value between -2/3 and 1.5:
$6(0)^2<6+5(0)$
$0<6+0$
$0<6 \rightarrow \text{ TRUE}$
Third test with a value more than 1.5:
$6(2)^2<6+5(2)$
$6(4)<6+10$
$24<16 \rightarrow \text{ FALSE}$
These tests show that the inequality $6x^2<6+5x$ is valid for values between -2/3 and 1.5 (not including them) i.e. $(-\frac{2}{3},1.5)$
