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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