Solve each of the following quadratic equations:

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=3 \frac{1}{3}, \quad x \neq 5,7$

$\Rightarrow \frac{(x-4)(x-7)+(x-5)(x-6)}{(x-5)(x-7)}=\frac{10}{3}$

$\Rightarrow \frac{x^{2}-11 x+28+x^{2}-11 x+30}{x^{2}-12 x+35}=\frac{10}{3}$

$\Rightarrow \frac{2 x^{2}-22 x+58}{x^{2}-12 x+35}=\frac{10}{3}$

$\Rightarrow \frac{x^{2}-11 x+29}{x^{2}-12 x+35}=\frac{5}{3}$

$\Rightarrow 3 x^{2}-33 x+87=5 x^{2}-60 x+175$

$\Rightarrow 2 x^{2}-27 x+88=0$

$\Rightarrow 2 x^{2}-16 x-11 x+88=0$

$\Rightarrow 2 x(x-8)-11(x-8)=0$

$\Rightarrow(x-8)(2 x-11)=0$

$\Rightarrow x-8=0$ or $2 x-11=0$

$\Rightarrow x=8$ or $x=\frac{11}{2}$

Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.