(2x+10)+(6x+28)=(10x-12)

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Solution for (2x+10)+(6x+28)=(10x-12) equation: 



(2x+10)+(6x+28)=(10x-12)

We move all terms to the left:

(2x+10)+(6x+28)-((10x-12))=0

We get rid of parentheses

2x+6x-((10x-12))+10+28=0



We calculate terms in parentheses: -((10x-12)), so:

(10x-12)

We get rid of parentheses

10x-12

Back to the equation:

-(10x-12)



We add all the numbers together, and all the variables

8x-(10x-12)+38=0

We get rid of parentheses

8x-10x+12+38=0

We add all the numbers together, and all the variables

-2x+50=0

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

-2x=-50

x=-50/-2

x=+25

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