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

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



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

We move all terms to the left:

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



Domain of the equation: 2+x^2)!=0

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

x^2)!=-2

x!=-2/1

x!=-2

x∈R



We use the square of the difference formula

-(3(x-1)/2+x^2)+x^2-9=0

We multiply all the terms by the denominator


-(3(x-1)+x^2*2+x^2)+x^2)-9*2=0



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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses

x^2+x^2*2+3x-3

Wy multiply elements

x^2+2x^2+3x-3

We add all the numbers together, and all the variables

3x^2+3x-3

Back to the equation:

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



We add all the numbers together, and all the variables

x^2)-9*2-(3x^2+3x=0

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