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x^{2}-2x+1=12
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x^{2}-2x+1-12=12-12
Subtract 12 from both sides of the equation.
x^{2}-2x+1-12=0
Subtracting 12 from itself leaves 0.
x^{2}-2x-11=0
Subtract 12 from 1.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-11\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\left(-11\right)}}{2}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+44}}{2}
Multiply -4 times -11.
x=\frac{-\left(-2\right)±\sqrt{48}}{2}
Add 4 to 44.
x=\frac{-\left(-2\right)±4\sqrt{3}}{2}
Take the square root of 48.
x=\frac{2±4\sqrt{3}}{2}
The opposite of -2 is 2.
x=\frac{4\sqrt{3}+2}{2}
Now solve the equation x=\frac{2±4\sqrt{3}}{2} when ± is plus. Add 2 to 4\sqrt{3}.
x=2\sqrt{3}+1
Divide 2+4\sqrt{3} by 2.
x=\frac{2-4\sqrt{3}}{2}
Now solve the equation x=\frac{2±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from 2.
x=1-2\sqrt{3}
Divide 2-4\sqrt{3} by 2.
x=2\sqrt{3}+1 x=1-2\sqrt{3}
The equation is now solved.
x^{2}-2x+1=12
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\left(x-1\right)^{2}=12
Factor x^{2}-2x+1. 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-1\right)^{2}}=\sqrt{12}
Take the square root of both sides of the equation.
x-1=2\sqrt{3} x-1=-2\sqrt{3}
Simplify.
x=2\sqrt{3}+1 x=1-2\sqrt{3}
Add 1 to both sides of the equation.