2x+2x^2-(1-2x)+x^2+3=5x^2+2-(2x^2-4)+2x

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Solution for 2x+2x^2-(1-2x)+x^2+3=5x^2+2-(2x^2-4)+2x equation: 



2x+2x^2-(1-2x)+x^2+3=5x^2+2-(2x^2-4)+2x

We move all terms to the left:

2x+2x^2-(1-2x)+x^2+3-(5x^2+2-(2x^2-4)+2x)=0

We add all the numbers together, and all the variables

2x^2+x^2+2x-(-2x+1)-(5x^2+2-(2x^2-4)+2x)+3=0

We add all the numbers together, and all the variables

3x^2+2x-(-2x+1)-(5x^2+2-(2x^2-4)+2x)+3=0

We get rid of parentheses

3x^2+2x+2x-(5x^2+2-(2x^2-4)+2x)-1+3=0



We calculate terms in parentheses: -(5x^2+2-(2x^2-4)+2x), so:

5x^2+2-(2x^2-4)+2x

determiningTheFunctionDomain
5x^2-(2x^2-4)+2x+2

We add all the numbers together, and all the variables

5x^2+2x-(2x^2-4)+2

We get rid of parentheses

5x^2-2x^2+2x+4+2

We add all the numbers together, and all the variables

3x^2+2x+6

Back to the equation:

-(3x^2+2x+6)



We add all the numbers together, and all the variables

3x^2+4x-(3x^2+2x+6)+2=0

We get rid of parentheses

3x^2-3x^2+4x-2x-6+2=0

We add all the numbers together, and all the variables

2x-4=0

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

2x=4

x=4/2

x=2

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