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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

4(d-4)-(3(d+1))+2=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-17=0

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

d=17

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