Solve for x
x=1
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4-4x+x^{2}=x^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-x\right)^{2}.
4-4x+x^{2}-x^{2}=0
Subtract x^{2} from both sides.
4-4x=0
Combine x^{2} and -x^{2} to get 0.
-4x=-4
Subtract 4 from both sides. Anything subtracted from zero gives its negation.
x=\frac{-4}{-4}
Divide both sides by -4.
x=1
Divide -4 by -4 to get 1.
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