Answer     $2, \frac{3}{4}$

Because it is clear that $x\ne 0$, let's divide the numerators and denominators by $x$: $$\frac{2}{2x - 5 +\frac{3}{x}} + \frac{13}{2x + 1 +\frac{3}{x}}=6$$

Let $y=2x +\frac{3}{x} -2$, then the above relation becomes $$\frac{2}{y-3}+\frac{13}{y+3}=6\implies \frac{15y-33}{y^2-9}=6\implies y= -1, \frac{7}{2}$$

Setting this back to $y=2x+\frac{3}{x}-2$ finds only two real solutions $x =\boxed{2}, \boxed{\frac{3}{4}}$.