How do you factor ${{x}^{2}}-6x-4$?

Hint: In the problem we have a quadratic equation of the variable $x$. So, we will compare the given equation with the normal form of the quadratic equation which is $a{{x}^{2}}+bx+c$ and calculate the values of $a$, $b$, $c$. Now we have the roots of the quadratic equation $a{{x}^{2}}+bx+c$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. From this formula we will calculate the roots of the given equation. If $\alpha $, $\beta $ are the roots of the given equation, then we can write ${{x}^{2}}-6x-4=\left( x-\alpha \right)\left( x-\beta \right)$ where $x-\alpha $, $x-\beta $ are the roots of the given equation.

Complete step by step answer:
Given equation, ${{x}^{2}}-6x-4$.
Comparing the above equation with $a{{x}^{2}}+bx+c$, then the values of $a$, $b$, $c$are
$a=1$, $b=-6$, $c=-4$.
Now the roots of the given equation are
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above formula, then we will get
$x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 1 \right)\left( -4 \right)}}{2\left( 1 \right)}$
When we multiply a negative sign with the negative sign then we will get positive sign, so
$\begin{align}
  & x=\dfrac{6\pm \sqrt{36+16}}{2} \\
 & \Rightarrow x=\dfrac{6\pm \sqrt{52}}{2} \\
\end{align}$
We can write $52=4\times 13$, then
$x=\dfrac{6\pm \sqrt{4\times 13}}{2}$
We know that $\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}$, then we will get
\[\Rightarrow x=\dfrac{6\pm \sqrt{4}\times \sqrt{13}}{2}\]
We have $\sqrt{4}=2$, then we will get
\[\Rightarrow x=\dfrac{6\pm 2\sqrt{13}}{2}\]
Taking $2$ common from the numerator, then we will get
\[\begin{align}
  & \Rightarrow x=\dfrac{2\left( 3\pm \sqrt{13} \right)}{2} \\
 & \Rightarrow x=3\pm \sqrt{13} \\
\end{align}\]
$\therefore $The roots of the given equation are $x=3+\sqrt{13}$ and $x=3-\sqrt{13}$.
So, we can write
$\begin{align}
  & {{x}^{2}}-6x-4=\left( x-\left( 3+\sqrt{13} \right) \right)\left( x-\left( 3-\sqrt{13} \right) \right) \\
 & \Rightarrow {{x}^{2}}-6x-4=\left( x-3-\sqrt{13} \right)\left( x-3+\sqrt{13} \right) \\
\end{align}$

Hence the factors of the given equation are $x-3-\sqrt{13}$, $x-3+\sqrt{13}$.

Note: In the problem they mentioned to calculate the factors, so we have calculated the roots of the given equation, from the roots we calculated the factors. If they have asked to find roots or solve the equation, then no need to write the factors calculating roots is enough.