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2=x^{2}-6x+9+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.
2=x^{2}-6x+10
Add 9 and 1 to get 10.
x^{2}-6x+10=2
Swap sides so that all variable terms are on the left hand side.
x^{2}-6x+10-2=0
Subtract 2 from both sides.
x^{2}-6x+8=0
Subtract 2 from 10 to get 8.
a+b=-6 ab=8
To solve the equation, factor x^{2}-6x+8 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.
-1,-8 -2,-4
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.
-1-8=-9 -2-4=-6
Calculate the sum for each pair.
a=-4 b=-2
The solution is the pair that gives sum -6.
\left(x-4\right)\left(x-2\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=4 x=2
To find equation solutions, solve x-4=0 and x-2=0.
2=x^{2}-6x+9+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.
2=x^{2}-6x+10
Add 9 and 1 to get 10.
x^{2}-6x+10=2
Swap sides so that all variable terms are on the left hand side.
x^{2}-6x+10-2=0
Subtract 2 from both sides.
x^{2}-6x+8=0
Subtract 2 from 10 to get 8.
a+b=-6 ab=1\times 8=8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.
-1,-8 -2,-4
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.
-1-8=-9 -2-4=-6
Calculate the sum for each pair.
a=-4 b=-2
The solution is the pair that gives sum -6.
\left(x^{2}-4x\right)+\left(-2x+8\right)
Rewrite x^{2}-6x+8 as \left(x^{2}-4x\right)+\left(-2x+8\right).
x\left(x-4\right)-2\left(x-4\right)
Factor out x in the first and -2 in the second group.
\left(x-4\right)\left(x-2\right)
Factor out common term x-4 by using distributive property.
x=4 x=2
To find equation solutions, solve x-4=0 and x-2=0.
2=x^{2}-6x+9+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.
2=x^{2}-6x+10
Add 9 and 1 to get 10.
x^{2}-6x+10=2
Swap sides so that all variable terms are on the left hand side.
x^{2}-6x+10-2=0
Subtract 2 from both sides.
x^{2}-6x+8=0
Subtract 2 from 10 to get 8.
x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36-32}}{2}
Multiply -4 times 8.
x=\frac{-\left(-6\right)±\sqrt{4}}{2}
Add 36 to -32.
x=\frac{-\left(-6\right)±2}{2}
Take the square root of 4.
x=\frac{6±2}{2}
The opposite of -6 is 6.
x=\frac{8}{2}
Now solve the equation x=\frac{6±2}{2} when ± is plus. Add 6 to 2.
x=4
Divide 8 by 2.
x=\frac{4}{2}
Now solve the equation x=\frac{6±2}{2} when ± is minus. Subtract 2 from 6.
x=2
Divide 4 by 2.
x=4 x=2
The equation is now solved.
2=x^{2}-6x+9+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.
2=x^{2}-6x+10
Add 9 and 1 to get 10.
x^{2}-6x+10=2
Swap sides so that all variable terms are on the left hand side.
x^{2}-6x=2-10
Subtract 10 from both sides.
x^{2}-6x=-8
Subtract 10 from 2 to get -8.
x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-6x+9=-8+9
Square -3.
x^{2}-6x+9=1
Add -8 to 9.
\left(x-3\right)^{2}=1
Factor x^{2}-6x+9. 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-3\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
x-3=1 x-3=-1
Simplify.
x=4 x=2
Add 3 to both sides of the equation.