3(3-c)-29=21c-4(5+6c)

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Solution for 3(3-c)-29=21c-4(5+6c) equation: 



3(3-c)-29=21c-4(5+6c)

We move all terms to the left:

3(3-c)-29-(21c-4(5+6c))=0

We add all the numbers together, and all the variables

3(-1c+3)-(21c-4(6c+5))-29=0

We multiply parentheses

-3c-(21c-4(6c+5))+9-29=0



We calculate terms in parentheses: -(21c-4(6c+5)), so:

21c-4(6c+5)

We multiply parentheses

21c-24c-20

We add all the numbers together, and all the variables

-3c-20

Back to the equation:

-(-3c-20)



We add all the numbers together, and all the variables

-3c-(-3c-20)-20=0

We get rid of parentheses

-3c+3c+20-20=0

We add all the numbers together, and all the variables

=0

c=0/1

c=0

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