2x^2+40=(x+4)(2x-7)

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Solution for 2x^2+40=(x+4)(2x-7) equation:



2x^2+40=(x+4)(2x-7)
We move all terms to the left:
2x^2+40-((x+4)(2x-7))=0
We multiply parentheses ..
2x^2-((+2x^2-7x+8x-28))+40=0
We calculate terms in parentheses: -((+2x^2-7x+8x-28)), so:
(+2x^2-7x+8x-28)
We get rid of parentheses
2x^2-7x+8x-28
We add all the numbers together, and all the variables
2x^2+x-28
Back to the equation:
-(2x^2+x-28)
We get rid of parentheses
2x^2-2x^2-x+28+40=0
We add all the numbers together, and all the variables
-1x+68=0
We move all terms containing x to the left, all other terms to the right
-x=-68
x=-68/-1
x=+68

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