9c(c-11)+10(5c-3)=3c(c+15)+c(6c-3)-30

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Solution for 9c(c-11)+10(5c-3)=3c(c+15)+c(6c-3)-30 equation: 



9c(c-11)+10(5c-3)=3c(c+15)+c(6c-3)-30

We move all terms to the left:

9c(c-11)+10(5c-3)-(3c(c+15)+c(6c-3)-30)=0

We multiply parentheses

9c^2-99c+50c-(3c(c+15)+c(6c-3)-30)-30=0



We calculate terms in parentheses: -(3c(c+15)+c(6c-3)-30), so:

3c(c+15)+c(6c-3)-30

We multiply parentheses

3c^2+6c^2+45c-3c-30

We add all the numbers together, and all the variables

9c^2+42c-30

Back to the equation:

-(9c^2+42c-30)



We add all the numbers together, and all the variables

9c^2-49c-(9c^2+42c-30)-30=0

We get rid of parentheses

9c^2-9c^2-49c-42c+30-30=0

We add all the numbers together, and all the variables

-91c=0

c=0/-91

c=0

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