Solve the following quadratic equations by factorization:

We have been given

$\frac{1}{x}-\frac{1}{x-2}=3$

$-2=3 x^{2}-6 x$

$3 x^{2}-6 x+2=0$

$3 x^{2}-(3+\sqrt{3}) x-(3-\sqrt{3}) x+3-\sqrt{3}+\sqrt{3}-1=0$

$x(3 x-3-\sqrt{3})+\left(\frac{-3+\sqrt{3}}{3}\right)(3 x-3-\sqrt{3})=0$

$\left(\frac{3 x-3+\sqrt{3}}{3}\right)(3 x-3-\sqrt{3})=0$

$(\sqrt{3} x-\sqrt{3}+1)(\sqrt{3} x-\sqrt{3}-1)=0$

Therefore,

$\sqrt{3} x-\sqrt{3}+1=0$

$\sqrt{3} x=\sqrt{3}-1$

$x=\frac{\sqrt{3}-1}{\sqrt{3}}$

or,

$\sqrt{3} x-\sqrt{3}-1=0$

$\sqrt{3} x=\sqrt{3}+1$

$x=\frac{\sqrt{3}+1}{\sqrt{3}}$

Hence, $x=\frac{\sqrt{3}-1}{\sqrt{3}}$ or $x=\frac{\sqrt{3}+1}{\sqrt{3}}$.