Solve the following quadratic equations by factorization:

We have been given

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

$x^{2}-1=3 x$

$x^{2}-3 x-1=0$

$x^{2}-\left(\frac{3-\sqrt{13}}{2}\right) x-\left(\frac{3+\sqrt{13}}{2}\right) x+\left(\frac{3+\sqrt{13}}{2}\right)\left(\frac{3-\sqrt{13}}{2}\right)=0$

$x\left(x-\left(\frac{3-\sqrt{13}}{2}\right)\right)-\left(\frac{3+\sqrt{13}}{2}\right)\left(x-\left(\frac{3-\sqrt{13}}{2}\right)\right)=0$

$\left[x-\left(\frac{3-\sqrt{13}}{2}\right)\right]\left[x-\left(\frac{3+\sqrt{13}}{2}\right)\right]=0$

Therefore,

$x-\left(\frac{3-\sqrt{13}}{2}\right)=0$

$x=\frac{3-\sqrt{13}}{2}$

or,

$x-\left(\frac{3+\sqrt{13}}{2}\right)=0$

$x=\frac{3+\sqrt{13}}{2}$

Hence, $x=\frac{3-\sqrt{13}}{2}$ or $x=\frac{3+\sqrt{13}}{2}$.