We must factorise the denominator to identify the linear factors first:

$x^2+x-6=(x+3)(x-2)$

It follows that we must find $A$ and $B$ such that

$\frac{5-3x}{x^2+x-6}=\frac{A}{x+3}+\frac{B}{x-2}=\frac{A(x-2)+B(x+3)}{(x+3)(x-2)}$

It follows that $5-3x=A(x-2)+B(x+3)=(A+B)x-2A+3B$ That is, $A+B=-3$ and $-2A+3B=5$. Solving gives $A=-\frac{14}{5}$ and $=-\frac{1}{5}$. Hence,

$\frac{5-3x}{x^2+x-6}=-\frac{\frac{14}{5}}{x+3}-\frac{\frac{1}{5}}{x-2}=-\frac{14}{5(x+3)}-\frac{1}{5(x-2)}$