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

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Solution for 3(d-4)+(4-d)2=16 equation: 



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3d-2d-12+8-16=0

We add all the numbers together, and all the variables

d-20=0

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

d=20

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