#### Answer

The values of x are $1+\sqrt{2}$ and $1-\sqrt{2}$.

#### Work Step by Step

Considered the equation,
${{x}^{2}}-2x-1=0$ ,
Now, follow the steps to solve the equation,
1. Divide both sides by $a$ to make the leading coefficient 1.
${{x}^{2}}-2x-1=0$
Here, the leading coefficient is already 1.
2. Isolate the variable terms on one side of the equation and constant terms on the other side.
${{x}^{2}}-2x=1$
3. Add the square of half the coefficient of x to both sides of the equation.
$\begin{align}
& {{x}^{2}}-2x+{{\left( \frac{2}{2} \right)}^{2}}=1+{{\left( \frac{2}{2} \right)}^{2}} \\
& {{x}^{2}}-2x+{{\left( 1 \right)}^{1}}=1+{{\left( 1 \right)}^{1}} \\
& {{x}^{2}}-2x+1=1+1 \\
& {{x}^{2}}-2x+1=2
\end{align}$
Factor the resulting perfect square trinomial,
$\begin{align}
& {{x}^{2}}-2x+1=2 \\
& {{\left( x-1 \right)}^{2}}=2 \\
\end{align}$
4. Use the square root property and solve for x.
Take the square root of both sides,
$\begin{align}
& \sqrt{{{\left( x-1 \right)}^{2}}}=\sqrt{2} \\
& x-1=\pm \sqrt{2} \\
& x=1\pm \sqrt{2}
\end{align}$
Hence, the values of x are $1+\sqrt{2}$ and $1-\sqrt{2}$.