2(2+3k)=2(k-6)

Solution for 2(2+3k)=2(k-6) equation:

2(2+3k)=2(k-6)

We move all terms to the left:

2(2+3k)-(2(k-6))=0

We add all the numbers together, and all the variables

2(3k+2)-(2(k-6))=0

We multiply parentheses

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


We calculate terms in parentheses: -(2(k-6)), so:

2(k-6)

We multiply parentheses

2k-12

Back to the equation:

-(2k-12)


We get rid of parentheses

6k-2k+12+4=0

We add all the numbers together, and all the variables

4k+16=0

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

4k=-16

k=-16/4

k=-4