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

Given :

$x^{2}-4 x-1=0$

On comparing it with $a x^{2}+b x+c=0$, we get :

$a=1, b=-4$ and $c=-1$

Discriminant $D$ is given by :

$D=\left(b^{2}-4 a c\right)$

$=(-4)^{2}-4 \times 1 \times(-1)$

$=16+4$

$=20$

$=20>0$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by :

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-4)+\sqrt{20}}{2 \times 1}=\frac{4+2 \sqrt{5}}{2}=\frac{2(2+\sqrt{5})}{2}=(2+\sqrt{5})$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-4)-\sqrt{20}}{2}=\frac{4-2 \sqrt{5}}{2}=\frac{2(2-\sqrt{5})}{2}=(2-\sqrt{5})$

Thus, the roots of the equation are $(2+\sqrt{5})$ and $(2-\sqrt{5})$.