8x^2+34x+21=(2x+7)(4x+3)

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Solution for 8x^2+34x+21=(2x+7)(4x+3) equation:



8x^2+34x+21=(2x+7)(4x+3)
We move all terms to the left:
8x^2+34x+21-((2x+7)(4x+3))=0
We multiply parentheses ..
8x^2-((+8x^2+6x+28x+21))+34x+21=0
We calculate terms in parentheses: -((+8x^2+6x+28x+21)), so:
(+8x^2+6x+28x+21)
We get rid of parentheses
8x^2+6x+28x+21
We add all the numbers together, and all the variables
8x^2+34x+21
Back to the equation:
-(8x^2+34x+21)
We add all the numbers together, and all the variables
8x^2+34x-(8x^2+34x+21)+21=0
We get rid of parentheses
8x^2-8x^2+34x-34x-21+21=0
We add all the numbers together, and all the variables
=0
x=0/1
x=0

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