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

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



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

We move all terms to the left:

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

We use the square of the difference formula

x^2+(2x-3)^2-(3x^2-2)+25=0

We get rid of parentheses

x^2-3x^2+(2x-3)^2+2+25=0

We add all the numbers together, and all the variables

-2x^2+(2x-3)^2+27=0

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

-2x^2+(2x-3)^2=-27

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