Absolute value equation infinite solutions $$|3-x|+4x=5|2+x|-13$$
One of the solutions is $[3,\infty)$
I'm not familiar with interval solutions for absolute equations. 
How to solve for this interval?

The first thing to do is to notice that if $x$ is large and positive, then
$|3-x|$ is the same as $x-3$, so the equation becomes
$$
(x-3)+4x = 5(x+2) -13 \\
5x-3 = 5x -3
$$ 
and so any time $x$ is that large, the equation is automatically true.  How large is "that large"?
Well, as long as $x\geq 3$ the essential property that $|3-x| = x-3$ holds. So the equation is autmatically satisfied for $x\geq 3$, that is, on the intervale $[3,+\infty)$.
(By the way, the square brace on the left means $x$ is greater than or equal to $3$; the parenthesis on the right means $x < \infty$. 
If $x<3$ then $|3-x| = 3-x$ and if $x \geq -2$  the equation becomes
$$
(3-x)+4x = 5(x+2)-13\\
3x+3 = 5x -3 \\
x=3
$$
and that adds nothing new.  But if $x<-2$ the equation becomes
$$
(3-x)+4x = -5(x+2)-13\\
3x+3 = -5x -23 \\
8x=-26 \\
x= -\frac{13}{4}
$$
and that is an isolated additional solution:
$$
|3-(-\frac{13}{4})| +4(-\frac{13}{4})= \frac{25}{4}-13= -\frac{27}{4}\\
5|2+(-\frac{13}{4})|-13 = 5(\frac54)-13 = -\frac{27}{4}
$$
