(2n+1.2)+(2n+1.2)+2n+1.2)=(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)

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Solution for (2n+1.2)+(2n+1.2)+2n+1.2)=(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4) equation: 



(2n+1.2)+(2n+1.2)+2n+1.2)=(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)

We move all terms to the left:

(2n+1.2)+(2n+1.2)+2n+1.2)-((3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4))=0

We add all the numbers together, and all the variables

2n+(2n+1.2)+(2n+1.2)+1.2)-((3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4)+(3n-4))=0

We get rid of parentheses

2n+2n+2n+1.2)-((3n-4)+3n+3n+3n+3n+3n+1.2+1.2-4-4-4-4-4)=0

We add all the numbers together, and all the variables

21n+1.2)-((3n-4)=0

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