Compute the sum of all the roots of
$(2x+3)(x-4)+(2x+3)(x-6)=0$
$\textbf{(A) } \frac{7}{2}\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 7\qquad \textbf{(E) } 13$
We expand to get $2x^2-8x+3x-12+2x^2-12x+3x-18=0$ which is $4x^2-14x-30=0$ after combining like terms. Using the quadratic part of [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas), we find the sum of the roots is $\frac{14}4 = \boxed{\textbf{(A) }7/2}$.