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

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



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

We move all terms to the left:

(x-2)(x+2)+3-(1-(x+3)+x^2)=0

We use the square of the difference formula

-(1-(x+3)+x^2)+x^2-4+3=0



We calculate terms in parentheses: -(1-(x+3)+x^2), so:

1-(x+3)+x^2

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

We get rid of parentheses

x^2-x-3+1

We add all the numbers together, and all the variables

x^2-1x-2

Back to the equation:

-(x^2-1x-2)



We add all the numbers together, and all the variables

x^2-(x^2-1x-2)-1=0

We get rid of parentheses

x^2-x^2+1x+2-1=0

We add all the numbers together, and all the variables

x+1=0

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

x=-1

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