Finding the roots?

$\begin{aligned} 3^{2x^2-7x+3} & = 4^{x^2-x-6} \\ 3^{(2x-1)(x-3)} & = 4^{(x+2)(x-3)} \\ (2x-1)(x-3)\log 3 & = (x+2)(x-3) \log 4 \end{aligned}$

Therefore, $x=3$ is a root. And the other root is given by:

$\begin{aligned} (2x-1)\log 3 & = (x+2) \log 4 \\ & = 2(x+2)\log 2 \\ 2x\log 3 - 2x \log 2 & = \log 3 + 4 \log 2 \\ \implies x & = \frac {\log 3 + 4 \log 2}{2\log 3 - 2 \log 2} \\ & = \frac {1 + 4 \cdot \frac {\log 2}{\log 3}}{2 - \frac {\log 4}{\log 3}} \end{aligned}$

Therefore, $A+B+C+D+E+F+G = 1+4+2+3+2+4+3 = \boxed{19}$

$\begin{aligned}3^{2x^2-7x+3}&=4^{x^2-x-6}\\\iff 2x^2-7x+3&=(x^2-x-6)\log_34\\\iff (2-\log_34)x^2-(7-\log_34)x+3+6\log_34&=0\\\iff (2-k)x^2-(7-k)x+(3+6k)&=0\end{aligned}$

Where $k:=\log_34$ .

Knowing $x=3$ is a root, we can use polynomial long division (dividing by $x-3$ ) to see that this holds when $(x-3)[(2-k)x-(1+2k)]=0$ .

The roots are then $x=3$ or $x=\dfrac{1+2k}{2-k}=\dfrac{1+2\frac{\log 4}{\log 3}}{2-\frac{\log 4}{\log 3}}$ , since $k=\log_34=\dfrac{\log 4}{\log 3}$ (where the logarithm has the base of any positive number not equal to $1$ ).

The answer is then $1+2+4+3+2+4+3=\color{#20A900}{\boxed{19}}$ .

One may also apply the quadratic formula here to get the same result, however this is tricky and errors may occur. If doing so, I would suggest making a substitution like I did for $k$ to reduce the chance of making mistakes.