jmc

By the quadratic formula, the roots of the equation are $$x=\frac{-(11)\pm \sqrt{(11{)}^2-4(\frac{3}{2})c}}{2(\frac{3}{2})},$$which simplifies to $$x=\frac{-11\pm \sqrt{121-6c}}{3}.$$This looks exactly like our target, except that we have to get the $121-6c$ under the square root to equal $7$. So, we solve the equation $121-6c=7$, which yields $c=\boxed{19}$.