3(d+I)=2(3D-9)

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Solution for 3(d+I)=2(3D-9) equation: 



3(d+)=2(3-9)

We move all terms to the left:

3(d+)-(2(3-9))=0

We add all the numbers together, and all the variables

3(+d)-(2(-6))=0

We add all the numbers together, and all the variables

3(+d)+12=0

We multiply parentheses

3d+12=0

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

3d=-12

d=-12/3

d=-4

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