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.

Let's look at the question:

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

Now to answer this, we need to know what the domain of a function is. Without getting into technical details, we can think of the domain as the values of $x$ that give us values of $y$. When answering a question to find the domain of a function, it is often more useful to try to find values of $x$ that do not give any value of $y$.
There are three basic kinds of functions to pay attention to: rational (fractions), sqaure roots, and logarithms. Here you have the first two. (I suspect that you haven't learned about logarithms, so we will ignore those for today.) For rational functions, the denominator cannot be zero. For square root functions, the radicand (stuff under the sqaure root symbol) cannot be negative. We can use these two facts to answer this problem.
First we set the denominator to zero:
$$\sqrt{x^2+x-2}=0$$
You can square both sides of this equation and use the quadratic formula (or factor or complete the square) to find that $x = 1$ or $x = -2$ are solutions to the equation. This means that these two values of $x$ are not in the domain.
The next step is to determine which values of $x$ make the denominator of the original function negative. You can use the solutions to the above step to determine this. To do this, we first note that $x^2+x-2$ is zero when $x = 1$ or $x = -2$. So we know that every other value of $x$ will make the quadratic expression either positive or negative. So let's look at values of $x$ less than $-2$, between $-2$ and $1$ and greater than $1$. Which of these "intervals" make our quadratic function positive?
