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

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



2(X+1)^2+3(X+2)(X-2)+X^2=(2X+1)(3X-2)-1
We move all terms to the left:
2(X+1)^2+3(X+2)(X-2)+X^2-((2X+1)(3X-2)-1)=0
We use the square of the difference formula
X^2+X^2+2(X+1)^2-((2X+1)(3X-2)-1)-4=0
We multiply parentheses ..
X^2+X^2-((+6X^2-4X+3X-2)-1)+2(X+1)^2-4=0
We calculate terms in parentheses: -((+6X^2-4X+3X-2)-1), so:
(+6X^2-4X+3X-2)-1
We get rid of parentheses
6X^2-4X+3X-2-1
We add all the numbers together, and all the variables
6X^2-1X-3
Back to the equation:
-(6X^2-1X-3)
We add all the numbers together, and all the variables
2X^2-(6X^2-1X-3)+2(X+1)^2-4=0
We get rid of parentheses
2X^2-6X^2+1X+2(X+1)^2+3-4=0
We add all the numbers together, and all the variables
-4X^2+X+2(X+1)^2-1=0
We move all terms containing X to the left, all other terms to the right
-4X^2+X+2(X+1)^2=1

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