3(2+c)=2(c-5)=

Solution for 3(2+c)=2(c-5)= equation:

3(2+c)=2(c-5)=

We move all terms to the left:

3(2+c)-(2(c-5))=0

We add all the numbers together, and all the variables

3(c+2)-(2(c-5))=0

We multiply parentheses

3c-(2(c-5))+6=0


We calculate terms in parentheses: -(2(c-5)), so:

2(c-5)

We multiply parentheses

2c-10

Back to the equation:

-(2c-10)


We get rid of parentheses

3c-2c+10+6=0

We add all the numbers together, and all the variables

c+16=0

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

c=-16