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x^{2}-2x+1-2x=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.
x^{2}-4x+1=0
Combine -2x and -2x to get -4x.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4}}{2}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{12}}{2}
Add 16 to -4.
x=\frac{-\left(-4\right)±2\sqrt{3}}{2}
Take the square root of 12.
x=\frac{4±2\sqrt{3}}{2}
The opposite of -4 is 4.
x=\frac{2\sqrt{3}+4}{2}
Now solve the equation x=\frac{4±2\sqrt{3}}{2} when ± is plus. Add 4 to 2\sqrt{3}.
x=\sqrt{3}+2
Divide 4+2\sqrt{3} by 2.
x=\frac{4-2\sqrt{3}}{2}
Now solve the equation x=\frac{4±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 4.
x=2-\sqrt{3}
Divide 4-2\sqrt{3} by 2.
x=\sqrt{3}+2 x=2-\sqrt{3}
The equation is now solved.
x^{2}-2x+1-2x=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.
x^{2}-4x+1=0
Combine -2x and -2x to get -4x.
x^{2}-4x=-1
Subtract 1 from both sides. Anything subtracted from zero gives its negation.
x^{2}-4x+\left(-2\right)^{2}=-1+\left(-2\right)^{2}
Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-4x+4=-1+4
Square -2.
x^{2}-4x+4=3
Add -1 to 4.
\left(x-2\right)^{2}=3
Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-2\right)^{2}}=\sqrt{3}
Take the square root of both sides of the equation.
x-2=\sqrt{3} x-2=-\sqrt{3}
Simplify.
x=\sqrt{3}+2 x=2-\sqrt{3}
Add 2 to both sides of the equation.