Solve each of the following quadratic equations:

$\frac{3 x-4}{7}+\frac{7}{3 x-4}=\frac{5}{2}, \quad x \neq \frac{4}{3}$

$\Rightarrow \frac{(3 x-4)^{2}+49}{7(3 x-4)}=\frac{5}{2}$

$\Rightarrow \frac{9 x^{2}-24 x+16+49}{21 x-28}=\frac{5}{2}$

$\Rightarrow \frac{9 x^{2}-24 x+65}{21 x-28}=\frac{5}{2}$

$\Rightarrow 18 x^{2}-48 x+130=105 x-140$

$\Rightarrow 18 x^{2}-153 x+270=0$

$\Rightarrow 2 x^{2}-17 x+30=0$

$\Rightarrow 2 x^{2}-12 x-5 x+30=0$

$\Rightarrow 2 x(x-6)-5(x-6)=0$

$\Rightarrow(x-6)(2 x-5)=0$

$\Rightarrow x-6=0$ or $2 x-5=0$

$\Rightarrow x=6$ or $x=\frac{5}{2}$

Hence, 6 and $\frac{5}{2}$ are the roots of the given equation.