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:

$\begin{array}{l}
D=\left(b^{2}-4 a c\right) \\
=(-4)^{2}-4 \times 1 \times(-1) \\
=16+4 \\
=20 \\
=20>0
\end{array}$

Hence, the roots of the equation are real.

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

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