Answer
$=\displaystyle \frac{2x^{2}+5x+12}{(x-3)(x+2)}, \quad x\neq-2,3$
Work Step by Step
The denominators are different.
Find a common denominator.
LCD=$(x+2)(x-3)$
$\displaystyle \frac{3x}{x-3}-\frac{x+4}{x+2}= \displaystyle \frac{3x(x-3)}{(x+2)(x-3)}-\frac{(x+4)(x+2)}{(x+2)(x-3)}$
$=\displaystyle \frac{3x^{2}+6x-(x^{2}+x-12)}{(x-3)(x+2)}$
$=\displaystyle \frac{3x^{2}+6x-x^{2}-x+12}{(x-3)(x+2)}$
$=\displaystyle \frac{2x^{2}+5x+12}{(x-3)(x+2)}$
... exclude values that yield 0 in the denominator:
$x\neq-2,3$
... and, cancel common factors
to factor $2x^{2}+5x+12$, we search for factors of ac=24 whose sum is b=5 and here we can't find any.
So here, there are no common factors to cancel.
$=\displaystyle \frac{2x^{2}+5x+12}{(x-3)(x+2)}, \quad x\neq-2,3$
