(x-7)^2+2x=(2x-1)(x-2)

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Solution for (x-7)^2+2x=(2x-1)(x-2) equation:



(x-7)^2+2x=(2x-1)(x-2)
We move all terms to the left:
(x-7)^2+2x-((2x-1)(x-2))=0
We add all the numbers together, and all the variables
2x+(x-7)^2-((2x-1)(x-2))=0
We multiply parentheses ..
-((+2x^2-4x-1x+2))+2x+(x-7)^2=0
We calculate terms in parentheses: -((+2x^2-4x-1x+2)), so:
(+2x^2-4x-1x+2)
We get rid of parentheses
2x^2-4x-1x+2
We add all the numbers together, and all the variables
2x^2-5x+2
Back to the equation:
-(2x^2-5x+2)
We add all the numbers together, and all the variables
2x-(2x^2-5x+2)+(x-7)^2=0
We get rid of parentheses
-2x^2+2x+5x+(x-7)^2-2=0
We add all the numbers together, and all the variables
-2x^2+7x+(x-7)^2-2=0
We move all terms containing x to the left, all other terms to the right
-2x^2+7x+(x-7)^2=2

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