Find the domain of $f(x)=\frac{3x+1}{\sqrt{x^2+x-2}}$ 
Find the domain of $f(x)=\dfrac{3x+1}{\sqrt{x^2+x-2}}$

This is my work so far:
$$\dfrac{3x+1}{\sqrt{x^2+x-2}}\cdot \sqrt{\dfrac{x^2+x-2}{x^2+x-2}}$$
$$\dfrac{(3x+1)(\sqrt{x^2+x-2})}{x^2+x-2}$$
$(3x+1)(\sqrt{x^2+x-2})$ = $\alpha$ (Just because it's too much to type)
$$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{1-4(1)(-2)}}{2}\right]}$$
$$\dfrac{\alpha}{\left[\dfrac{-1\pm \sqrt{9}}{2}\right]}$$
$$\dfrac{\alpha}{\left[\left(\dfrac{-1+3}{2}\right)\left(\dfrac{-1-3}{2}\right)\right]}$$
$$\dfrac{\alpha}{(1)(-2)}$$
Now, I checked on WolframAlpha and the domain is $x\in \mathbb R: x\lt -2$ or $x\gt 1$
But my question is, what do I do with the top of the problem? Or does it just not matter at all.

Sometimes (like this one!) things get pretty clearer when we make algebra and geometry interplay. This kind of questions are attacked around here in senior year of junior high school (advanced mathematics) without any calculus at all, of course.
First, you must know that $\,x^2+x-2=(x+2)(x-1)\,$ is a parabola and, in fact, a "smiling" one, meaning: it opens upwards, and its roots or zeros are $\,x_{1,2}=-2,1\,$ (and for this we need either the quadratic formula or factoring the trinomial).
So we have a smiling parabola, and from the geometric picture we get at once this parabola in non-positive  iff $\,-2\leq x\leq 1\,$ , so that it is positive whenever $\,x<-2\,\,\,or\,\,\,x>1\,$.
