(3x+1)4=(2x+8)+(x+8)+(4x+2)

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Solution for (3x+1)4=(2x+8)+(x+8)+(4x+2) equation: 



(3x+1)4=(2x+8)+(x+8)+(4x+2)

We move all terms to the left:

(3x+1)4-((2x+8)+(x+8)+(4x+2))=0

We multiply parentheses

12x-((2x+8)+(x+8)+(4x+2))+4=0



We calculate terms in parentheses: -((2x+8)+(x+8)+(4x+2)), so:

(2x+8)+(x+8)+(4x+2)

We get rid of parentheses

2x+x+4x+8+8+2

We add all the numbers together, and all the variables

7x+18

Back to the equation:

-(7x+18)



We get rid of parentheses

12x-7x-18+4=0

We add all the numbers together, and all the variables

5x-14=0

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

5x=14

x=14/5

x=2+4/5

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