(x-2)(x+2)+3x=1-(x+3)+x2

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Solution for (x-2)(x+2)+3x=1-(x+3)+x2 equation: 



(x-2)(x+2)+3x=1-(x+3)+x2

We move all terms to the left:

(x-2)(x+2)+3x-(1-(x+3)+x2)=0

We add all the numbers together, and all the variables

3x+(x-2)(x+2)-(1-(x+3)+x2)=0

We use the square of the difference formula

x^2+3x-(1-(x+3)+x2)-4=0



We calculate terms in parentheses: -(1-(x+3)+x2), so:

1-(x+3)+x2

determiningTheFunctionDomain
-(x+3)+x2+1

We add all the numbers together, and all the variables

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 get rid of parentheses

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

We add all the numbers together, and all the variables

4x-2=0

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

4x=2

x=2/4

x=1/2

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