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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



We calculate terms in parentheses: -((3x^2+2x-4)), so:

(3x^2+2x-4)

We get rid of parentheses

3x^2+2x-4

Back to the equation:

-(3x^2+2x-4)



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

=0

x=0/1

x=0

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