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

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Solution for 6(d+2)-4+d=3(d-3) equation: 



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

d+6(d+2)-(3(d-3))-4=0

We multiply parentheses

d+6d-(3(d-3))+12-4=0



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

3(d-3)

We multiply parentheses

3d-9

Back to the equation:

-(3d-9)



We add all the numbers together, and all the variables

7d-(3d-9)+8=0

We get rid of parentheses

7d-3d+9+8=0

We add all the numbers together, and all the variables

4d+17=0

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

4d=-17

d=-17/4

d=-4+1/4

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