-2(x2+x-1)+5(x+1)(x+1)=3(x+1)(x-2)

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Solution for -2(x2+x-1)+5(x+1)(x+1)=3(x+1)(x-2) equation: 



-2(x2+x-1)+5(x+1)(x+1)=3(x+1)(x-2)

We move all terms to the left:

-2(x2+x-1)+5(x+1)(x+1)-(3(x+1)(x-2))=0

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

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



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

3(x+1)(x-2)

We multiply parentheses ..

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

We multiply parentheses

3x^2-6x+3x-6

We add all the numbers together, and all the variables

3x^2-3x-6

Back to the equation:

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



We multiply parentheses

-2x^2+5x^2+5x+5x-2x-(3x^2-3x-6)+5+2=0

We get rid of parentheses

-2x^2+5x^2-3x^2+5x+5x-2x+3x+6+5+2=0

We add all the numbers together, and all the variables

11x+13=0

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

11x=-13

x=-13/11

x=-1+2/11

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