I-g+2(3+g)=-4(g+1)

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Solution for I-g+2(3+g)=-4(g+1) equation: 



-I+2(3+I)=-4(I+1)

We move all terms to the left:

-I+2(3+I)-(-4(I+1))=0

We add all the numbers together, and all the variables

-I+2(I+3)-(-4(I+1))=0

We add all the numbers together, and all the variables

-1I+2(I+3)-(-4(I+1))=0

We multiply parentheses

-1I+2I-(-4(I+1))+6=0



We calculate terms in parentheses: -(-4(I+1)), so:

-4(I+1)

We multiply parentheses

-4I-4

Back to the equation:

-(-4I-4)



We add all the numbers together, and all the variables

I-(-4I-4)+6=0

We get rid of parentheses

I+4I+4+6=0

We add all the numbers together, and all the variables

5I+10=0

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

5I=-10

I=-10/5

I=-2

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