If this is the equation x = 2 + (1/x) then first multiply through by x to get:
x2 = 2x + 1
Rearrange:
x2 - 2x - 1 = 0
Solve using the quadratic formula:
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.414\: \!213\: \!562\: \!373\: \!095}}\\
{\mathtt{x}} = {\mathtt{2.414\: \!213\: \!562\: \!373\: \!095}}\\
\end{array} \right\}$$
This doesn't make sense. How can x = something that has x.
If x = 1: 1 = (2+1)/1 ? NO & 1 = 2 + (1/1) ? NO.
Also, parentheses are very important! Is it (2+1)/x or 2+(1/x)?
If this is the equation x = 2 + (1/x) then first multiply through by x to get:
x2 = 2x + 1
Rearrange:
x2 - 2x - 1 = 0
Solve using the quadratic formula:
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.414\: \!213\: \!562\: \!373\: \!095}}\\
{\mathtt{x}} = {\mathtt{2.414\: \!213\: \!562\: \!373\: \!095}}\\
\end{array} \right\}$$