Answer
$\left\{ 7,-2,6,-1 \right\}$.
Work Step by Step
${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
Let $u=\left( {{x}^{2}}-5x-1 \right)$ and ${{u}^{2}}={{\left( {{x}^{2}}-5x-1 \right)}^{2}}$. Substitute these values in the expression ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
$\begin{align}
& {{u}^{2}}-18u+65=0 \\
& {{u}^{2}}-13u-5u+65=0 \\
& u\left( u-13 \right)-5\left( u-13 \right)=0 \\
& \left( u-13 \right)\left( u-5 \right)=0
\end{align}$
Thus,
$u=5$ or $u=13$.
Now, we find:
$\begin{align}
& \left( {{x}^{2}}-5x-1 \right)=5 \\
& {{x}^{2}}-5x-1-5=0 \\
& {{x}^{2}}-5x-6=0 \\
& {{x}^{2}}-6x+x-6=0
\end{align}$
$\begin{align}
& {{x}^{2}}-6x+x-6=0 \\
& x\left( x-6 \right)+1\left( x-6 \right)=0 \\
& \left( x-6 \right)\left( x+1 \right)=0
\end{align}$
So,
$\begin{align}
& \left( x+1 \right)=0 \\
& x=-1
\end{align}$
And,
$\begin{align}
& \left( x-6 \right)=0 \\
& x=6
\end{align}$
Therefore, the values of x are 6 or $-1$.
Now, replacing u with $\left( {{x}^{2}}-5x-1 \right)$,
$\begin{align}
& \left( {{x}^{2}}-5x-1 \right)=13 \\
& {{x}^{2}}-5x-1-13=0 \\
& {{x}^{2}}-5x-14=0 \\
& {{x}^{2}}-7x+2x-14=0
\end{align}$
$\begin{align}
& {{x}^{2}}-7x+2x-14=0 \\
& x\left( x-7 \right)+2\left( x-7 \right)=0 \\
& \left( x-7 \right)\left( x+2 \right)=0
\end{align}$
$\begin{align}
& \left( x-7 \right)=0 \\
& x=7
\end{align}$
$\begin{align}
& \left( x+2 \right)=0 \\
& x=-2
\end{align}$
Thus, the values of x are 7 or $-2$.
Check:
Put $x=6$ in the expression ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
$\begin{align}
{{\left( {{6}^{2}}-5\cdot 6-1 \right)}^{2}}-18\left( {{6}^{2}}-5\cdot 6-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
{{\left( 36-30-1 \right)}^{2}}-18\left( 36-30-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
\left( 25 \right)-18\left( 5 \right)+65\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
Hence, $x=6$ is true for ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
Put $x=-1$ in the expression ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
$\begin{align}
{{\left( {{\left( -1 \right)}^{2}}-5\cdot \left( -1 \right)-1 \right)}^{2}}-18\left( {{\left( -1 \right)}^{2}}-5\cdot \left( -1 \right)-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
{{\left( 1+5-1 \right)}^{2}}-18\left( 1+5-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
\left( 25 \right)-18\left( 5 \right)+65\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
Hence, $x=-1$ is true for ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
Put $x=7$ in the expression ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
$\begin{align}
{{\left( {{7}^{2}}-5\cdot 7-1 \right)}^{2}}-18\left( {{7}^{2}}-5\cdot 7-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
{{\left( 49-35-1 \right)}^{2}}-18\left( 49-35-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
169-234+65\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
Hence, $x=7$ is true for ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
Put $x=-2$ in the expression ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
$\begin{align}
{{\left[ {{\left( -2 \right)}^{2}}-5\cdot \left( -2 \right)-1 \right]}^{2}}-18\left( {{\left( -2 \right)}^{2}}-5\cdot \left( -2 \right)-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
{{\left( 4+10-1 \right)}^{2}}-18\left( 4+10-1 \right)+65\overset{?}{\mathop{=}}\,0 & \\
169-234+65\overset{?}{\mathop{=}}\,0 & \\
0=0 & \\
\end{align}$
Hence, $x=-2$ is true for ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$.
Thus, the solutions of the expression ${{\left( {{x}^{2}}-5x-1 \right)}^{2}}-18\left( {{x}^{2}}-5x-1 \right)+65=0$ are $\left\{ 7,-2,6,-1 \right\}$.
