2x(x-3)(x+1)+(2x-3)^2=(x+1)^2+2x^2-x^2

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Solution for 2x(x-3)(x+1)+(2x-3)^2=(x+1)^2+2x^2-x^2 equation:



2x(x-3)(x+1)+(2x-3)^2=(x+1)^2+2x^2-x^2
We move all terms to the left:
2x(x-3)(x+1)+(2x-3)^2-((x+1)^2+2x^2-x^2)=0
We multiply parentheses ..
-((x+1)^2+2x^2-x^2)+2x(+x^2+x-3x-3)+(2x-3)^2=0
We calculate terms in parentheses: -((x+1)^2+2x^2-x^2), so:
(x+1)^2+2x^2-x^2
determiningTheFunctionDomain 2x^2-x^2+(x+1)^2
We add all the numbers together, and all the variables
x^2+(x+1)^2
Back to the equation:
-(x^2+(x+1)^2)
We multiply parentheses
2x^3+2x^2-6x^2-6x-(x^2+(x+1)^2)+(2x-3)^2=0
We calculate terms in parentheses: -(x^2+(x+1)^2), so:
x^2+(x+1)^2
We do not support expression: x^3

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