(12-2y)/3)^2-y+2y^2=12

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Solution for (12-2y)/3)^2-y+2y^2=12 equation: 



(12-2y)/3)^2-y+2y^2=12

We move all terms to the left:

(12-2y)/3)^2-y+2y^2-(12)=0



Domain of the equation: 3)^2-y+2y^2-12!=0

y!=0/1

y!=0

y∈R



We add all the numbers together, and all the variables

(-2y+12)/3)^2-y+2y^2-12=0

We multiply all the terms by the denominator


(-2y+12)=0

We get rid of parentheses

-2y+12=0

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

-2y=-12

y=-12/-2

y=+6

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