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

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



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

We move all terms to the left:

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

We use the square of the difference formula

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

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

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

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