2(d-3)-6=4(d+2)+10

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Solution for 2(d-3)-6=4(d+2)+10 equation: 



2(d-3)-6=4(d+2)+10

We move all terms to the left:

2(d-3)-6-(4(d+2)+10)=0

We multiply parentheses

2d-(4(d+2)+10)-6-6=0



We calculate terms in parentheses: -(4(d+2)+10), so:

4(d+2)+10

We multiply parentheses

4d+8+10

We add all the numbers together, and all the variables

4d+18

Back to the equation:

-(4d+18)



We add all the numbers together, and all the variables

2d-(4d+18)-12=0

We get rid of parentheses

2d-4d-18-12=0

We add all the numbers together, and all the variables

-2d-30=0

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

-2d=30

d=30/-2

d=-15

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