Solve the following quadratic equations by factorization:

We have been given

$\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}$

$\frac{2 x(x-3)+(2 x-5)(x-4)}{x^{2}-7 x+12}=\frac{25}{3}$

$6 x^{2}-18 x+6 x^{2}-24 x-15 x+60=25 x^{2}-175 x+300$

$13 x^{2}-118 x+240=0$

$13 x^{2}-78 x-40 x+240=0$

$13 x(x-6)-40(x-6)=0$

$(x-6)(13 x-40)=0$

Therefore,

$x-6=0$

$x=6$

or,

$13 x-40=0$

$13 x=40$

$x=\frac{40}{13}$

Hence, $x=6$ or $x=\frac{40}{13}$.