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x^{2}-2x+1-4=0
Subtract 4 from both sides.
x^{2}-2x-3=0
Subtract 4 from 1 to get -3.
a+b=-2 ab=-3
To solve the equation, factor x^{2}-2x-3 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
a=-3 b=1
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
\left(x-3\right)\left(x+1\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=3 x=-1
To find equation solutions, solve x-3=0 and x+1=0.
x^{2}-2x+1-4=0
Subtract 4 from both sides.
x^{2}-2x-3=0
Subtract 4 from 1 to get -3.
a+b=-2 ab=1\left(-3\right)=-3
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
a=-3 b=1
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.
\left(x^{2}-3x\right)+\left(x-3\right)
Rewrite x^{2}-2x-3 as \left(x^{2}-3x\right)+\left(x-3\right).
x\left(x-3\right)+x-3
Factor out x in x^{2}-3x.
\left(x-3\right)\left(x+1\right)
Factor out common term x-3 by using distributive property.
x=3 x=-1
To find equation solutions, solve x-3=0 and x+1=0.
x^{2}-2x+1=4
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+1-4=4-4
Subtract 4 from both sides of the equation.
x^{2}-2x+1-4=0
Subtracting 4 from itself leaves 0.
x^{2}-2x-3=0
Subtract 4 from 1.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)}}{2}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4+12}}{2}
Multiply -4 times -3.
x=\frac{-\left(-2\right)±\sqrt{16}}{2}
Add 4 to 12.
x=\frac{-\left(-2\right)±4}{2}
Take the square root of 16.
x=\frac{2±4}{2}
The opposite of -2 is 2.
x=\frac{6}{2}
Now solve the equation x=\frac{2±4}{2} when ± is plus. Add 2 to 4.
x=3
Divide 6 by 2.
x=-\frac{2}{2}
Now solve the equation x=\frac{2±4}{2} when ± is minus. Subtract 4 from 2.
x=-1
Divide -2 by 2.
x=3 x=-1
The equation is now solved.
x^{2}-2x+1=4
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}=4
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{4}
Take the square root of both sides of the equation.
x-1=2 x-1=-2
Simplify.
x=3 x=-1
Add 1 to both sides of the equation.