jmc

We complete the square: $$\begin{array}{rl}6j^2-4j+12& =6\left(j^2-\frac{2}{3}j\right)+12\\ & =6\left(j^2-\frac{2}{3}j+\frac{1}{9}\right)+12-\frac{6}{9}\\ & =6{\left(j-\frac{1}{3}\right)}^2+\frac{34}{3}\end{array}$$Then $q=\frac{34}{3}$ and $p=-\frac{1}{3}$. The question asks for $\frac{q}{p}$, which is equal to $\boxed{-34}$.