Factor $2x^3 + 3x^2 - 2x$ I am having the most difficult time factoring this equation out. I should have paid more attention in precalculus in high school :(
$2x^3 + 3x^2 - 2x = x(2x^2 + 3x - 2) = x(2x-1)(x+2)$
How do you get the $(2x-1)(x+2)$ out of $(2x^2+3x-2)$?
Quadratic equation didn't work, I can't take out a common variable, I am out of options in my head

You can always use the quadratic equation to factor a quadratic polynomial $Q(x)$.
If the solutions to $Q(x)=0$ are $x=r$ and $x=s$, then the polynomial can be factored as
$$Q(x)=k(x-r)(x-s)$$
where $k$ is the leading coefficient of $Q(x)$.
In your case, $Q(x)=2x^2+3x-2$ and the quadratic formula says the solutions to $Q(x)=0$ are
$$x=\frac{-3 \pm\sqrt{3^2-4(2)(-2)}}{2(2)}=\frac{-3 \pm 5}{4}$$
So the solutions are $x=\frac12$ and $x=-2$.
This means $Q(x)$ factors as $2(x-\frac12)(x+2)$. It is convenient to combine the leading constant with the first factor to get rid of the fraction, giving
$$\boxed{Q(x)=(2x-1)(x+2)}.$$

Addendum:
It may be worth remembering how the quadratic formula is derived in the first place. Believe it or not, it's through factoring (!). Here's how. Suppose we have a quadratic equation (assume $a\neq0$, otherwise it's really just a linear equation):
$$ax^2+bx+c=0$$
We work towards the goal of building the constant discriminant $b^2-4ac$ on the right side: 
$$ax^2+bx = \boxed{-c}\tag{subtract $c$ from both sides}$$
$$4a^2x^2 + 4abx = \boxed{-4ac}\tag{multiply both sides by $4a$}$$
$$4a^2x^2 + 4abx +b^2= \boxed{b^2-4ac}\tag{add $b^2$ to both sides}$$
When you do this, the left side is always a perfect square polynomial (the square of a binomial). You can then factor the left side:
$$\overbrace{(2ax+b)(2ax+b)}^{4a^2x^2 + 4abx +b^2\textrm{ factored}}=
\overbrace{b^2-4ac}^{\textrm{constant discriminant}}$$
Assuming that $b^2-4ac$ isn't negative, we can write $b^2-4ac=(\sqrt{b^2-4ac})^2$ so
$$(2ax+b)^2 = (\sqrt{b^2-4ac})^2$$
$$(2ax+b)^2 - (\sqrt{b^2-4ac})^2 = 0$$
which can be factored, being a difference of squares, as
$$(2ax+b - \sqrt{b^2-4ac})(2ax+b + \sqrt{b^2-4ac})=0$$
Dividing both sides by $2a$ twice, and then multiplying by $a$, we have
$$a\left(x+\frac{b- \sqrt{b^2-4ac}}{2a}\right)\left(x+\frac{b+ \sqrt{b^2-4ac}}{2a}\right)=0$$
$$a\left(x-\left[\frac{-b+ \sqrt{b^2-4ac}}{2a}\right]\right)\left(x-\left[\frac{-b- \sqrt{b^2-4ac}}{2a}\right]\right)=0$$
The left side is indeed of the form $k(x-r)(x-s)$, where $k$ is the leading coefficient of $Q(x)$ (that is, $a$) and $r,s$ are the solutions $\tfrac{-b\pm \sqrt{b^2-4ac}}{2a}$from the quadratic formula.
If the discriminant is negative, the same formula works but the square root will produce an imaginary number. The quadratic will still factor, but the factors will contain complex numbers. In other words, if $b^2-4ac<0$, we can write
$$\pm\sqrt{b^2-4ac} = \pm i\sqrt{-(b^2-4ac)}.$$
