6=-2(2+c)-(4+2c)

Solution for 6=-2(2+c)-(4+2c) equation:

6=-2(2+c)-(4+2c)

We move all terms to the left:

6-(-2(2+c)-(4+2c))=0

We add all the numbers together, and all the variables

-(-2(c+2)-(2c+4))+6=0


We calculate terms in parentheses: -(-2(c+2)-(2c+4)), so:

-2(c+2)-(2c+4)

We multiply parentheses

-2c-(2c+4)-4

We get rid of parentheses

-2c-2c-4-4

We add all the numbers together, and all the variables

-4c-8

Back to the equation:

-(-4c-8)


We get rid of parentheses

4c+8+6=0

We add all the numbers together, and all the variables

4c+14=0

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

4c=-14

c=-14/4

c=-3+1/2