#### Answer

The value of \[{{x}^{2}}-x\]is \[2\].

#### Work Step by Step

The equation\[4\left( x-2 \right)+2=4x-2\left( 2-x \right)\].
Use the distributive property, \[a\left( b+c \right)=ab+ac\].
\[\begin{align}
& 4x-8+2=4x-4+2x \\
& 4x-6=6x-4 \\
\end{align}\]
Add \[4\]to both side,
\[\begin{align}
& 4x-6+4=6x-4+4 \\
& 4x-2=6x \\
\end{align}\]
Subtract \[4x\] from both side,
\[\begin{align}
& 4x-4x-2=6x-4x \\
& -2=2x
\end{align}\]
Divided by 2 both side,
\[\begin{align}
& \frac{-2}{2}=\frac{2x}{2} \\
& -1=x \\
& x=-1 \\
\end{align}\]
The solution set is \[\left\{ -1 \right\}\].
Check the proposed solution. Substitute \[-1\] for x in the original equation \[4\left( x-2 \right)+2=4x-2\left( 2-x \right)\].
\[\begin{align}
& 4\left( -1-2 \right)+2=4\left( -1 \right)-2\left( 2+1 \right) \\
& 4\left( -3 \right)+2=-4-2\left( 3 \right) \\
& -12+2=-4-6 \\
& -10=-10
\end{align}\]
This true statement \[-10=-10\] verifies that the solution set is \[\left\{ -1 \right\}\].
Now put \[x=-1\] in the expression \[{{x}^{2}}-x\].
\[\begin{align}
& {{x}^{2}}-x={{\left( -1 \right)}^{2}}-\left( -1 \right) \\
& =1+1 \\
& =2
\end{align}\]
Therefore, value of \[{{x}^{2}}-x\]is \[2\].