How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?

Depending on your level, you may not have seen the quadratic formula that JMoravitz and BLAZE posted. That formula is often the way to find the zeroes of and factor a quadratic.
But what Adam suggests may be more up your alley, and it's what I'll discuss as well: You can make educated guesses and then check to see if it works. The problem you have posted does not have many options to guess, so this method is fairly quick.
Consider your quadratic and what you expect the factored form to look like:
$$2x^2-x-3 = (\color{red}{\Box}x \color{red}{\pm} \color{red}{\Box})(\color{red}{\Box}x\color{red}{\pm}\color{red}{\Box})$$
You know that the factored form requires you to fill in those blanks, choosing the appropriate sign and number. What numbers can possibly satisfy the equation? 
Start with the $2x^2$ term on the left. This one is straightforward. In order to get $2x^2$, the coefficients of the $x$'s on the right hand side must be $2$ and $1$ if we are to have all integer coefficients.
$$2x^2-x-3 = (\color{green}{2}x \color{red}{\pm} \color{red}{\Box})(\color{green}{1}x\color{red}{\pm}\color{red}{\Box})$$
$$2x^2-x-3 = (2x \color{red}{\pm} \color{red}{\Box})(x\color{red}{\pm}\color{red}{\Box})$$
Now what about the $-3$ on the right? It's factors are $\pm1, \pm3$. We need to select two of these numbers and have them fill the remaining sections. You can try all combinations until you have the answer, but we can be smarter than that.
Since $-3$ is negative, one of these numbers will be positive and the other will be negative. Furthermore, the two factors will need to add to $-1$, after one of them is multiplied by $2$, since we have $-x$ on the left.
Notice that $2-3 = -1$. That's what we want. So the $2x$ multiplies a $1$ and the $x$ multiplies a $-3$.
$$2x^2-x-3 = (2x \color{green}{-3})(x\color{green}{+1})$$
We do a quick check:
$$(2x-3)(x+1) = 2x^2 +2x -3x -3 = 2x^2 -x -3$$
It works.
