<s>[INST] Solving for $x$ in an inequality $$\frac{x^2-3x-2}{x^2+5x+6}<\frac{2-x}{x^2-4}$$
First, I got the things it couldn't be, which were: -3, -2, and 2
I thought to factor everything I could, and I got
$$\frac{x^2-3x-2}{(x+3)(x+2)}<\frac{2-x}{(x+2)(x-2)}$$
I also cancelled out one value on the right side
$$\frac{x^2-3x-2}{(x+3)(x+2)}<\frac{-1}{x+2}$$
but now, I'm stuck.  
 [/INST] Starting from where you stopped, we have 
$$\dfrac{x^2-3x-2}{(x+3)(x+2)}+\dfrac{1}{x+2}<0$$
Adding, we have
$$\dfrac{(x-1)^2}{(x+3)(x+2)}<0$$
As the numerator is non-negative, we can ignore it for now.  So we need solutions to
$$(x+3)(x+2)<0$$
You should be able to do this now simply by splitting the number line into three regions, viz. $x < -3, -3 < x < -2, x > -2$ and checking in each.
Dont forget to exclude $x=1$ if it is in the feasible region!
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