jmc

When $x\in \left(-\frac{7}{3},2\right)$, we have $x\cdot (3x+1)-c<0$. This means that $x(3x+1)-c=0$ at $x=-\frac{7}{3}$ and $x=2$. We now know that $x(3x+1)-c=0$ is a quadratic equation with roots at $x=-\frac{7}{3}$ and $x=2$, and we want to use these roots to find a quadratic of the same form as the problem. $x=-\frac{7}{3}$ gives us $(3x+7)=0$ and $x=2$ gives us $(x-2)=0$. $$\begin{array}{rl}x(3x+1)-c& =(3x+7)(x-2)\\ & =3x^2+x-14\\ & =x(3x+1)-14.\end{array}$$ So, $c=\boxed{14}$.