3(4d+3)=12-(8d-5)+6d

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Solution for 3(4d+3)=12-(8d-5)+6d equation: 



3(4d+3)=12-(8d-5)+6d

We move all terms to the left:

3(4d+3)-(12-(8d-5)+6d)=0

We multiply parentheses

12d-(12-(8d-5)+6d)+9=0



We calculate terms in parentheses: -(12-(8d-5)+6d), so:

12-(8d-5)+6d

determiningTheFunctionDomain
-(8d-5)+6d+12

We add all the numbers together, and all the variables

6d-(8d-5)+12

We get rid of parentheses

6d-8d+5+12

We add all the numbers together, and all the variables

-2d+17

Back to the equation:

-(-2d+17)



We get rid of parentheses

12d+2d-17+9=0

We add all the numbers together, and all the variables

14d-8=0

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

14d=8

d=8/14

d=4/7

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