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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

2(k+2)-(3(-1k+6))=0

We multiply parentheses

2k-(3(-1k+6))+4=0



We calculate terms in parentheses: -(3(-1k+6)), so:

3(-1k+6)

We multiply parentheses

-3k+18

Back to the equation:

-(-3k+18)



We get rid of parentheses

2k+3k-18+4=0

We add all the numbers together, and all the variables

5k-14=0

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

5k=14

k=14/5

k=2+4/5

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