10=2(3d+4)+d-5

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Solution for 10=2(3d+4)+d-5 equation: 



10=2(3d+4)+d-5

We move all terms to the left:

10-(2(3d+4)+d-5)=0



We calculate terms in parentheses: -(2(3d+4)+d-5), so:

2(3d+4)+d-5

We add all the numbers together, and all the variables

d+2(3d+4)-5

We multiply parentheses

d+6d+8-5

We add all the numbers together, and all the variables

7d+3

Back to the equation:

-(7d+3)



We get rid of parentheses

-7d-3+10=0

We add all the numbers together, and all the variables

-7d+7=0

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

-7d=-7

d=-7/-7

d=1

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