<s>[INST] 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?
 [/INST] Hint: From here:
$$\frac{2x^2 - 43x - 45}{2x(x + \frac{3}{2})} \lt 0$$
you could say that the fraction is less than zero if and only if either (1) the numerator is positive and the denominator is negative or (2) the numerator is negative and the denominator is positive. So for case (2), you solve:
$$
2x^2 - 43x - 45 <0 \quad \text{and}\quad 2x(x +\frac{3}{2}) > 0.
$$
You might already know this, but the way you solve each of these two inequalities is by finding possible roots. So for example for the $2x(x+ \frac{3}{2}) > 0$, the roots are $x = 0$ and $x = -\frac{3}{2}$. Now you just need to find the (constant) signs of the expression $2x(x+\frac{3}{2})$ on the intervals $(-\infty, -\frac{3}{2})$, $(-\frac{3}{2}, 0)$, and $(0, \infty)$.
</s>