3(4d-7)+6=2(d+2)-1

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Solution for 3(4d-7)+6=2(d+2)-1 equation: 



3(4d-7)+6=2(d+2)-1

We move all terms to the left:

3(4d-7)+6-(2(d+2)-1)=0

We multiply parentheses

12d-(2(d+2)-1)-21+6=0



We calculate terms in parentheses: -(2(d+2)-1), so:

2(d+2)-1

We multiply parentheses

2d+4-1

We add all the numbers together, and all the variables

2d+3

Back to the equation:

-(2d+3)



We add all the numbers together, and all the variables

12d-(2d+3)-15=0

We get rid of parentheses

12d-2d-3-15=0

We add all the numbers together, and all the variables

10d-18=0

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

10d=18

d=18/10

d=1+4/5

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