I give up... Can someone please help/explain this to me?

Hi Lina,

Welcome to Web2.0calc forum  

This is called partial fractions, it comes in useful when doing calculus.

$\frac{3x+2}{x^2-x-12}=\frac{A}{(x-4)}+\frac{B}{(x+3)}\\ So\\ Ax+3A+Bx-4B=3x+2\\ (A+B)x+(3A-4B)=3x+2\\ A+B=3\qquad and \qquad 3A-4B=2\\ A=3-B\\ \text{sub to get B}\\ 3(3-B)-4B=2\\ -7B+9=2\\ B=1\\ A=3-1=2\\ \frac{3x+2}{x^2-x-12}=\frac{2}{(x-4)}+\frac{1}{(x+3)}\\ $

You can finish :)