Answer
$4\pm 2\sqrt{2} $
Work Step by Step
$f\left( x \right)=\frac{{{x}^{2}}}{x-2}+1$ and $g\left( x \right)=\frac{4x-2}{x-2}+\frac{x+4}{2}$
Using substitution:
$\begin{align}
& \frac{{{x}^{2}}}{x-2}+1=\frac{4x-2}{x-2}+\frac{x+4}{2} \\
& \frac{{{x}^{2}}}{x-2}-\frac{4x-2}{x-2}=\frac{x+4}{2}-1 \\
& \frac{{{x}^{2}}-4x+2}{x-2}=\frac{x+4-2}{2} \\
& 2\left( {{x}^{2}}-4x+2 \right)=\left( x-2 \right)\left( x+2 \right)
\end{align}$
Simplify:
$\begin{align}
& 2{{x}^{2}}-8x+4={{x}^{2}}-2x+2x-4 \\
& 2{{x}^{2}}-8x+4={{x}^{2}}-4 \\
& 2{{x}^{2}}-{{x}^{2}}-8x+4+4=0 \\
& {{x}^{2}}-8x+8=0
\end{align}$
Now compare the equation ${{x}^{2}}-8x+8=0 $with the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$,
Here,
$a=1,b=-8$and $c=8$
Substitute $a=1,b=-8$ and $c=8$ into the quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$,
$\begin{align}
& x=\frac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 1 \right)\left( 8 \right)}}{2\left( 1 \right)} \\
& =\frac{8\pm \sqrt{64-32}}{2} \\
& =\frac{8\pm \sqrt{32}}{2} \\
& =\frac{8\pm \sqrt{16\cdot 2}}{2}
\end{align}$
Simplify:
$\begin{align}
& x=\frac{8\pm 4\sqrt{2}}{2} \\
& =4\pm 2\sqrt{2}
\end{align}$
