<s>[INST] 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?
 [/INST] You can either use the quadratic formula for a general quadratic $ax^2+bx+c$ which is
$$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ which I will say no more of as JMoravitz has rigorously explained it. 
Or you can complete the square, or write
$$2x^2 \color{blue}{-x}-3$$
$$=2x^2+\color{blue}{2x-3x}-3$$
$$=2x(x+1)-3(x+1)$$
$$=\color{red}{(2x-3)(x+1)}$$
Notice, in the part marked $\color{blue}{\mathrm{blue}}$ I have simply rewritten $-x$ as $2x-3x$.
From then on you simply take out common factors. 
Note that this method is only valid when you have two numbers whose product is $-6$ and sum is $-1$. 
Or, put in another way for the general quadratic equation $ax^2 +bx +c$, this inspection method is only valid iff you can find two numbers whose product is $ac$ and sum is $b$. Otherwise using the quadratic formula is the best method as it suits all scenarios.  
For completing the square:
$$2x^2 -x-3$$
$$=2\left(x^2 -\frac{x}{2}-\frac{3}{2}\right)$$
$$=2\left(\left(x -\frac{1}{4}\right)^2-\frac{1}{16}-\frac{3}{2}\right)$$
$$=2\left(\left(x -\frac{1}{4}\right)^2-\left(\frac{5}{4}\right)^2\right)\tag{1}$$
$$=2\left(\left(x -\frac{1}{4}+\frac{5}{4}\right)-\left(x -\frac{1}{4}-\frac{5}{4}\right)\right)\tag{2}$$
$$=2\left((x+1)\left(x -\frac{3}{2}\right)\right)$$
$$=\color{red}{(2x-3)(x+1)}$$
as before. 
From $(1)$ to $(2)$ I used the difference of two squares formula such that $a^2-b^2=(a+b)(a-b) \space \space\forall a,b \in \mathbb{R}$
Hope this helps.
</s>