6(3c-5)+3=4c+20

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Solution for 6(3c-5)+3=4c+20 equation: 



6(3c-5)+3=4c+20

We move all terms to the left:

6(3c-5)+3-(4c+20)=0

We multiply parentheses

18c-(4c+20)-30+3=0

We get rid of parentheses

18c-4c-20-30+3=0

We add all the numbers together, and all the variables

14c-47=0

We move all terms containing c to the left, all other terms to the right

14c=47

c=47/14

c=3+5/14

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