The roots of the quadratic equation $3x^2+5x+k = 0$ are $\frac{-5\pm i\sqrt{131}}{6}$. What is $k$?
A quadratic with roots $\frac{-5\pm i\sqrt{131}}{6}$ has the form
\begin{align*}
3(x - \tfrac{-5+i\sqrt{131}}{6})(x - \tfrac{-5-i\sqrt{131}}{6}) = 0.
\end{align*}To find the value of $k$, we will expand out the product on the left-hand side.
By the difference of squares formula, we have
\begin{align*}
\left(x - \tfrac{-5+i\sqrt{131}}{6}\right) \left(x - \tfrac{-5-i\sqrt{131}}{6}\right) &= \left(\tfrac{-5+i\sqrt{131}}{6} - x\right) \left(\tfrac{-5-i\sqrt{131}}{6} - x\right) \\
&= \dfrac{(-5+i\sqrt{131})^2}{36} - \dfrac{(-5+i\sqrt{131})}{6}x - \dfrac{(-5-i\sqrt{131})}{6}x + x^2 \\
&= \dfrac{(131 - 10i \sqrt{131} + 25) - 6(-5+i\sqrt{131})x - 6(-5-i\sqrt{131})x}{36} \\
&= \dfrac{156 - 6(-5+i\sqrt{131})x - 6(-5-i\sqrt{131})x}{36} \\
&= \dfrac{156 - (-60+6i\sqrt{131})x - (-60-6i\sqrt{131})x}{36} \\
&= \dfrac{156 - 60x+6i\sqrt{131}x + 60x + 6i\sqrt{131}x}{36} \\
&= \dfrac{156 + 0 + 12i\sqrt{131}x}{36}.
\end{align*}Finally, we can conclude
$3(x - \tfrac{-5+i\sqrt{131}}{6})(x - \tfrac{-5-i\sqrt{131}}{6}) =3\cdot \dfrac{156 + 0 + 12i\sqrt{131}x}{36} = \dfrac{156+12i\sqrt{131}x}{12},$so $\dfrac{156+12i\sqrt{131}x}{12}=3x^2+5x+k$, which gives $\dfrac{156}{12}=k$. We find $k=\boxed{13}$.
To find the value of $k$, we can use the fact that in a quadratic equation of the form $ax^2+bx+c=0$, the value of $c$ is equal to the product of the roots.
In this case, the roots are ${\frac{-5+\sqrt{131} i}{6}}$ and ${\frac{-5-\sqrt{131} i}{6}}$. So, we can write:
$k = \left( {\frac{-5+\sqrt{131} i}{6}} \right) \left( {\frac{-5-\sqrt{131} i}{6}} \right)$
Let's simplify this expression step by step:
$k = \frac{1}{36} \left[ -5 \cdot (-5) + (-5) \cdot (-\sqrt{131} i) + (\sqrt{131} i) \cdot (-5) + (\sqrt{131} i) \cdot (-\sqrt{131} i) \right]$
$k = \frac{1}{36} \left[ 25 + 5 \sqrt{131} i - 5 \sqrt{131} i - 131 \right]$
$k = \frac{1}{36} \left[ -106 \right]$
$k = \boxed{-\frac{53}{18}}$