10(c-2)+4=3+c-19

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Solution for 10(c-2)+4=3+c-19 equation: 



10(c-2)+4=3+c-19

We move all terms to the left:

10(c-2)+4-(3+c-19)=0

We add all the numbers together, and all the variables

10(c-2)-(c-16)+4=0

We multiply parentheses

10c-(c-16)-20+4=0

We get rid of parentheses

10c-c+16-20+4=0

We add all the numbers together, and all the variables

9c=0

c=0/9

c=0

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