8-(20-4c)=4(c-3)

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Solution for 8-(20-4c)=4(c-3) equation: 



8-(20-4c)=4(c-3)

We move all terms to the left:

8-(20-4c)-(4(c-3))=0

We add all the numbers together, and all the variables

-(-4c+20)-(4(c-3))+8=0

We get rid of parentheses

4c-(4(c-3))-20+8=0



We calculate terms in parentheses: -(4(c-3)), so:

4(c-3)

We multiply parentheses

4c-12

Back to the equation:

-(4c-12)



We add all the numbers together, and all the variables

4c-(4c-12)-12=0

We get rid of parentheses

4c-4c+12-12=0

We add all the numbers together, and all the variables

=0

c=0/1

c=0

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