I14-2(3p+1)=6(4+p)

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Solution for I14-2(3p+1)=6(4+p) equation: 



14-2(3I+1)=6(4+I)

We move all terms to the left:

14-2(3I+1)-(6(4+I))=0

We add all the numbers together, and all the variables

-2(3I+1)-(6(I+4))+14=0

We multiply parentheses

-6I-(6(I+4))-2+14=0



We calculate terms in parentheses: -(6(I+4)), so:

6(I+4)

We multiply parentheses

6I+24

Back to the equation:

-(6I+24)



We add all the numbers together, and all the variables

-6I-(6I+24)+12=0

We get rid of parentheses

-6I-6I-24+12=0

We add all the numbers together, and all the variables

-12I-12=0

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

-12I=12

I=12/-12

I=-1

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