You have to deal with two separate failure modes. First, you cannot have a $0$ denominator, so $-3$ cannot be in the domain of the function. Secondly, $\frac{(2x-4)(5+x)}{x+3}$ must be non-negative, so you must solve the inequality $\frac{(2x-4)(5+x)}{x+3}\ge 0\;.$
The fraction is obviously positive when all three of the expressions $2x-4$, $5+x$, and $x+3$ are positive, but it’s also positive when exactly one of them is positive; in all other cases it is negative or zero. Now $2x-4>0$ when $x>2$, $5+x>0$ when $x>-5$, and $x+3>0$ when $x>-3$, so the relevant break points and intervals are as shown here:
---------|----------------------|-----------------------------------|------- -5 -3 2
When $x<-5$, none of the expressions is positive, so the fraction is negative; when $-5, only $5+x$ is positive, so the fraction is positive; when $-3, $5+x$ and $x+3$ are positive, so the fraction is negative; and when $x>2$, all three are positive, so the fraction is positive. Thus, the fraction is positive on $(-5,-3)\cup (2,\infty)\;.$ It is zero at $x=2$ and $x=-5$; neither of these makes the denominator zero, so they also belong in the domain, which is therefore $[-5,-3)\cup[2,\infty)\;.$