Answer
$y\in (-\infty, -5\cup(2,+\infty)$
Work Step by Step
We have to solve the inequality
$$y^2+3y>10.$$
Let's move all terms on one side
$$y^2+3y-10>0.$$
Factor the left side:
$$(y-2)(y+5)>0.$$
We determine the sign of the expression on the left side:
If $y\in(-\infty,-5)$, then $y-2<0$ and $y+5<0$, therefore $(y-2)(y+5)>0$.
If $y\in [-5,2]$, then $y-2\leq 0$ and $y+5\geq 0$, therefore $(y-2)(y+5)\leq 0$.
If $y\in (2,\infty)$, then $y-2>0$ and $y+5>0$, therefore $(y-2)(y+5)>0$.
Since we need the values of $y$ for which the expression is positive, the solution is $$y\in (-\infty, -5)\cup(2,+\infty)$$
