jmc

The problem simplifies slightly if we notice that $3\sqrt{5}=\sqrt{9\cdot 5}=\sqrt{45}$, and $2\sqrt{11}=\sqrt{4\cdot 11}=\sqrt{44}$. Writing the denominator this way, we have $$\frac{2}{\sqrt{45}+\sqrt{44}}=\frac{2}{\sqrt{45}+\sqrt{44}}\cdot \frac{\sqrt{45}-\sqrt{44}}{\sqrt{45}-\sqrt{44}}=2(\sqrt{45}-\sqrt{44}),$$since $45-44=1$ so the denominator is just 1. Rewriting what's left in simplest radical form again, we have $6\sqrt{5}-4\sqrt{11}$. Since $5<11$, we have $B=5$, and filling in the rest, $A=6$, $C=-4$, $D=11$, and $E=1$ (since there is no denominator, we just take it to be 1). Thus $A+B+C+D+E=\boxed{19}$.