jmc

Using the quadratic formula, we find that the roots of the quadratic are $\frac{-5\pm \sqrt{5^2-4(3)(k)}}{6}=\frac{-5\pm \sqrt{25-12k}}{6}$. Since the problem tells us that these roots must equal $\frac{-5\pm i\sqrt{11}}{6}$, we have $$\begin{array}{rl}\sqrt{25-12k}& =i\sqrt{11}\\ \Rightarrow \sqrt{25-12k}& =\sqrt{-11}\\ \Rightarrow 25-12k& =-11\\ \Rightarrow 12k& =36\\ \Rightarrow k& =\boxed{3}.\end{array}$$