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3-(x-5)^2=(2-x)(2+x)
We move all terms to the left:
3-(x-5)^2-((2-x)(2+x))=0
We add all the numbers together, and all the variables
-(x-5)^2-((-1x+2)(x+2))+3=0
We multiply parentheses ..
-((-1x^2-2x+2x+4))-(x-5)^2+3=0
We calculate terms in parentheses: -((-1x^2-2x+2x+4)), so:We get rid of parentheses
(-1x^2-2x+2x+4)
We get rid of parentheses
-1x^2-2x+2x+4
We add all the numbers together, and all the variables
-1x^2+4
Back to the equation:
-(-1x^2+4)
1x^2-(x-5)^2-4+3=0
We add all the numbers together, and all the variables
x^2-(x-5)^2-1=0
We move all terms containing x to the left, all other terms to the right
x^2-(x-5)^2=1
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