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

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



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

We move all terms to the left:

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

We multiply parentheses

4d-(3(d-1))-16+2=0



We calculate terms in parentheses: -(3(d-1)), so:

3(d-1)

We multiply parentheses

3d-3

Back to the equation:

-(3d-3)



We add all the numbers together, and all the variables

4d-(3d-3)-14=0

We get rid of parentheses

4d-3d+3-14=0

We add all the numbers together, and all the variables

d-11=0

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

d=11

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