Solve the following quadratic equations by factorization:

We have been given,

$7 x+\frac{3}{x}=35 \frac{3}{5}$

$7 x^{2}+3=\left(35+\frac{3}{5}\right) x$

$7 x^{2}-\left(35+\frac{3}{5}\right) x+3=0$

Therefore,

$7 x^{2}-35 x-\frac{3}{5} x+3=0$

$7 x(x-5)-\frac{3}{5}(x-5)=0$

$\left(7 x-\frac{3}{5}\right)(x-5)=0$

Therefore,

$7 x-\frac{3}{5}=0$

$7 x=\frac{3}{5}$

$x=\frac{3}{35}$

or,

$x-5=0$

$x=5$

Hence, $x=\frac{3}{35}$ or $x=5$.