Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$. 
But how do I get the last step? Or am I already dead wrong?

One approach is to find the (potential) "switchover" points. Note that one rational function can only become bigger than another after either intersecting or after a hole or vertical asymptote (of one or both). (Why?) There our no holes, and our asymptotes occur at $x=0$ and $x=-\frac32$. Intersection occurs when $x\neq 0$, $x\neq-\frac32$ and $$\frac{2x-13}{2x+3}=\frac{15}x$$ $$2x^2-13x=30x+45$$ $$2x^2-43x-45=0$$ $$2x^2-45x+2x-45=0$$ $$(2x-45)(x+1)=0.$$ Thus, our other possible switchover points are $x=-1$ and $x=\frac{45}2$.
Now, check the inequality on the intervals $\left(-\infty,-\frac32\right)$, $\left(-\frac32,-1\right)$, $(-1,0)$, $\left(0,\frac{45}2\right)$, and $\left(\frac{45}2,\infty\right)$ by using test points in each interval. Note that for the first interval a sign chart would readily tell us that the desired inequality fails (without need for a test point), since $\cfrac{2x-13}{2x+3}$ is positive and $\cfrac{15}x$ is negative.
