Let's say I had $\frac{x(x+a)}{x+b} \leq c$. How does one solve an equation like this?
What I did was split the answer up into two cases, one for $x+b>0$ and one for $x+b<0$. However, I had to further split these two cases into another two cases each, since I had something of the form $(x-b)(x+c) \geq 0$ for each original case (slightly different for each case). I'm not exactly sure if my method was correct and I can't find anything about the topic on the internet.
First solve $x(x+a)/(x+b)=c$, which (after multiplying by the denominator) is a quadratic. Those solutions and $-b$ (where the left side has a singularity), if they are distinct, divide the number line into three intervals, on each of which the difference between the sides is constant in sign, while crossing from one interval to the next should reverse the sign. So the inequality will be true in either the middle interval or the two outer ones.