n+(n+2)*3=(n+4)*14-58

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Solution for n+(n+2)*3=(n+4)*14-58 equation: 



n+(n+2)*3=(n+4)*14-58

We move all terms to the left:

n+(n+2)*3-((n+4)*14-58)=0

We multiply parentheses

n+3n-((n+4)*14-58)+6=0



We calculate terms in parentheses: -((n+4)*14-58), so:

(n+4)*14-58

We multiply parentheses

14n+56-58

We add all the numbers together, and all the variables

14n-2

Back to the equation:

-(14n-2)



We add all the numbers together, and all the variables

4n-(14n-2)+6=0

We get rid of parentheses

4n-14n+2+6=0

We add all the numbers together, and all the variables

-10n+8=0

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

-10n=-8

n=-8/-10

n=4/5

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