(x^2-2x+2)^2+3x(x^2+2x+2)=30x^2

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Solution for (x^2-2x+2)^2+3x(x^2+2x+2)=30x^2 equation: 



(x^2-2x+2)^2+3x(x^2+2x+2)=30x^2

We move all terms to the left:

(x^2-2x+2)^2+3x(x^2+2x+2)-(30x^2)=0

determiningTheFunctionDomain
-30x^2+(x^2-2x+2)^2+3x(x^2+2x+2)=0

We multiply parentheses

-30x^2+3x^3+6x^2+(x^2-2x+2)^2+6x=0

We do not support expression: x^3

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