Answer
$3a^2+a\gt10$
$3a^2+a-10\gt0$
$(3a-5)(a+2)\gt0$
$a\gt\frac{5}{3}$ and $a\lt-2$
Work Step by Step
Step 1: Subtract 10 from both sides of the inequality
Step 2: Factorize quadratic inequality (trinomial)
Step 3: The inequality function is greater than $0$ for $a\gt\frac{5}{3}$ and $a\lt-2$
