But how?

I separated the problem into cases that list all the possible values of $x_4$ : $x_4=0,1,2,3,4,5$ .

Then I used the stars and bars method on the resulting equation for each case:

$x_4=0$ : $x_1+x_2+x_3=20$ has $\frac{22!}{2!20!}=231$ solutions.

$x_4=1$ : $x_1+x_2+x_3=16$ has $\frac{18!}{2!16!}=153$ solutions.

$x_4=2$ : $x_1+x_2+x_3=12$ has $\frac{14!}{2!12!}=91$ solutions.

$x_4=3$ : $x_1+x_2+x_3=8$ has $\frac{10!}{2!8!}=45$ solutions.

$x_4=4$ : $x_1+x_2+x_3=4$ has $\frac{6!}{2!4!}=15$ solutions.

$x_4=5$ : $x_1+x_2+x_3=0$ has $\frac{2!}{2!0!}=1$ solution.

$231+153+91+45+15+1=536$ .