(7x-14)(3x+1)=(21x+7)(x-2)

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Solution for (7x-14)(3x+1)=(21x+7)(x-2) equation: 



(7x-14)(3x+1)=(21x+7)(x-2)

We move all terms to the left:

(7x-14)(3x+1)-((21x+7)(x-2))=0

We multiply parentheses ..

(+21x^2+7x-42x-14)-((21x+7)(x-2))=0



We calculate terms in parentheses: -((21x+7)(x-2)), so:

(21x+7)(x-2)

We multiply parentheses ..

(+21x^2-42x+7x-14)

We get rid of parentheses

21x^2-42x+7x-14

We add all the numbers together, and all the variables

21x^2-35x-14

Back to the equation:

-(21x^2-35x-14)



We get rid of parentheses

21x^2-21x^2+7x-42x+35x-14+14=0

We add all the numbers together, and all the variables

=0

x=0/1

x=0

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