Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

The given equation is

$x-\frac{1}{x}=3, x \neq 0$

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

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

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

This equation is of the form $a x^{2}+b x+c=0$, where $a=1, b=-3$ and $c=-1$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-3)^{2}-4 \times 1 \times(-1)=9+4=13>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{13}$

$\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-3)+\sqrt{13}}{2 \times 1}=\frac{3+\sqrt{13}}{2}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-3)-\sqrt{13}}{2 \times 1}=\frac{3-\sqrt{13}}{2}$

Hence, $\frac{3+\sqrt{13}}{2}$ and $\frac{3-\sqrt{13}}{2}$ are the roots of the given equation.