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

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Solution for 7(x-4)^2-3(x+5)^2=4(x+1)(x-2)-2 equation: 



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

We move all terms to the left:

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

We multiply parentheses ..

-(4(+x^2-2x+x-2)-2)+7(x-4)^2-3(x+5)^2=0



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

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

We multiply parentheses

4x^2-8x+4x-8-2

We add all the numbers together, and all the variables

4x^2-4x-10

Back to the equation:

-(4x^2-4x-10)



We get rid of parentheses

-4x^2+4x+7(x-4)^2-3(x+5)^2+10=0

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

-4x^2+4x+7(x-4)^2-3(x+5)^2=-10

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