21b(1+3b)=21b+63b^2

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Solution for 21b(1+3b)=21b+63b^2 equation: 



21b(1+3b)=21b+63b^2

We move all terms to the left:

21b(1+3b)-(21b+63b^2)=0

We add all the numbers together, and all the variables

-(21b+63b^2)+21b(3b+1)=0

We multiply parentheses

-(21b+63b^2)+63b^2+21b=0

We get rid of parentheses

-63b^2+63b^2-21b+21b=0

We add all the numbers together, and all the variables

=0

b=0/1

b=0

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