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2x^{2}-4x+2=0
Combine 5x and -9x to get -4x.
x^{2}-2x+1=0
Divide both sides by 2.
a+b=-2 ab=1\times 1=1
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.
a=-1 b=-1
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.
\left(x^{2}-x\right)+\left(-x+1\right)
Rewrite x^{2}-2x+1 as \left(x^{2}-x\right)+\left(-x+1\right).
x\left(x-1\right)-\left(x-1\right)
Factor out x in the first and -1 in the second group.
\left(x-1\right)\left(x-1\right)
Factor out common term x-1 by using distributive property.
\left(x-1\right)^{2}
Rewrite as a binomial square.
x=1
To find equation solution, solve x-1=0.
2x^{2}-4x+2=0
Combine 5x and -9x to get -4x.
x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\times 2}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 2}}{2\times 2}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-8\times 2}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-4\right)±\sqrt{16-16}}{2\times 2}
Multiply -8 times 2.
x=\frac{-\left(-4\right)±\sqrt{0}}{2\times 2}
Add 16 to -16.
x=-\frac{-4}{2\times 2}
Take the square root of 0.
x=\frac{4}{2\times 2}
The opposite of -4 is 4.
x=\frac{4}{4}
Multiply 2 times 2.
x=1
Divide 4 by 4.
2x^{2}-4x+2=0
Combine 5x and -9x to get -4x.
2x^{2}-4x=-2
Subtract 2 from both sides. Anything subtracted from zero gives its negation.
\frac{2x^{2}-4x}{2}=-\frac{2}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{4}{2}\right)x=-\frac{2}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-2x=-\frac{2}{2}
Divide -4 by 2.
x^{2}-2x=-1
Divide -2 by 2.
x^{2}-2x+1=-1+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=0
Add -1 to 1.
\left(x-1\right)^{2}=0
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{0}
Take the square root of both sides of the equation.
x-1=0 x-1=0
Simplify.
x=1 x=1
Add 1 to both sides of the equation.
x=1
The equation is now solved. Solutions are the same.