Turning Point

If we complete the square on the first parabola, we obtain $y = -x^2 + px + q - q^2 = -(x - p/2)^2 + (\frac{p^2}{4} - q^2 + q)$ with a vertex of $(\frac{p}{2}, \frac{p^2}{4} - q^2 + q).$ If we next substitute this point into the second parabola, one obtains:

$\frac{p^2}{4} - q^2 + q = 4(\frac{p}{2})^2 + 4(\frac{p}{2}) + \frac{19}{12}$ ;

or $3p^2 - 12q^2 + 12q = 12p^2 + 24p + 19;$

or $0 = 9p^2 + 24p + (12q^2 - 12q + 19);$

or $p = \frac{-24 \pm \sqrt{24^2 - 4(9)(12q^2 - 12q + 19)}}{18};$

or $p = \frac{-24 \pm 6\sqrt{16 - 12q^2 + 12q - 19}}{18};$

or $p = \frac{-4 \pm \sqrt{-12q^2 + 12q -3}}{3};$

or $p = \frac{-4 \pm \sqrt{-3(4q^2 - 4q +1)}}{3};$

or $p = \frac{-4 \pm \sqrt{-3(2q-1)^2}}{3};$

or $p = \frac{-4 \pm (2q-1)\sqrt{3}i}{3}.$

Since we require $p,q \in \mathbb{R}$ , this yields the pair $p = -\frac{4}{3}, q = \frac{1}{2}.$ Hence, $2q-3p = 2(\frac{1}{2}) - 3(-\frac{4}{3}) = 1 + 4 = \boxed{5}.$