jmc

We complete the square. First, we factor 3 out of the terms $3x^2+x$ to get $3\left(x^2+\frac{x}{3}\right)$. We can square $x+\frac{1}{6}$ to get $x^2+\frac{x}{3}+\frac{1}{36}$, so $$\begin{array}{rl}3\left(x^2+\frac{x}{3}\right)& =3\left[{\left(x+\frac{1}{6}\right)}^2-\frac{1}{36}\right]\\ & =3{\left(x+\frac{1}{6}\right)}^2-\frac{3}{36}\\ & =3{\left(x+\frac{1}{6}\right)}^2-\frac{1}{12},\end{array}$$and $$\begin{array}{rl}3\left(x^2+\frac{x}{3}\right)-4& =3{\left(x+\frac{1}{6}\right)}^2-\frac{1}{12}-4\\ & =3{\left(x+\frac{1}{6}\right)}^2-\frac{49}{12}.\end{array}$$We see that $k=\boxed{-\frac{49}{12}}$.