Let $x^2 + 6x + 4a = (x - p)(x - q) = 0$ and $x^2 + 2bx - 12 = (x - p)(x - r) = 0$ (where $x = p$ is the common solution). By Vieta's Formulae, we obtain the following relationships:
$p + q = -6$ AND $pq = 4a$ (i);
$p + r = -2b$ AND $pr = -12$ (ii)
If we solve for $a$ and $b$ in terms of the common root $p$, this results as:
$a = -(p^2 + 6p)/4, b = (12 - p^2)/(2p)$ (iii)
and substitution of these expressions into the latter two quadratics gives us:
$x^2 + 9x - (9/4)(p^2 + 6p) = 0$ AND $x^2 + (3/2p)(12 - p^2)x - 27 = 0$ (iv).
Let $x = t$ be a common solution to both quadratics in (iv) such that if we equate these two equations we obtain:
$9t - (9/4)(p^2 + 6p) = t^2 = (3/2p)(12 - p^2)t - 27$;
or $[9 + (3/2p)(p^2 -12)]t = (9/4)(p^2 + 6p) - 27$;
or $[(3p^2 + 18p - 36)/(2p)]t = (9p^2 + 54p - 108)/4$;
or $t = (9p^2 + 54p - 108)/4 * (2p)/(3p^2 + 18p - 36)$;
or $t = (p/2) * [9(p^2 + 6p -12)]/[3(p^2 + 6p -12)]$;
or $t = \frac{3p}{2}$.
QED