2(5c-17)-4=3(c+2)+26

Solution for 2(5c-17)-4=3(c+2)+26 equation:

2(5c-17)-4=3(c+2)+26

We move all terms to the left:

2(5c-17)-4-(3(c+2)+26)=0

We multiply parentheses

10c-(3(c+2)+26)-34-4=0


We calculate terms in parentheses: -(3(c+2)+26), so:

3(c+2)+26

We multiply parentheses

3c+6+26

We add all the numbers together, and all the variables

3c+32

Back to the equation:

-(3c+32)


We add all the numbers together, and all the variables

10c-(3c+32)-38=0

We get rid of parentheses

10c-3c-32-38=0

We add all the numbers together, and all the variables

7c-70=0

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

7c=70

c=70/7

c=10