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(2x+3)^2+(2x-3)^2=(8x+6)(x-1)+22
We move all terms to the left:
(2x+3)^2+(2x-3)^2-((8x+6)(x-1)+22)=0
We multiply parentheses ..
-((+8x^2-8x+6x-6)+22)+(2x+3)^2+(2x-3)^2=0
We calculate terms in parentheses: -((+8x^2-8x+6x-6)+22), so:We get rid of parentheses
(+8x^2-8x+6x-6)+22
We get rid of parentheses
8x^2-8x+6x-6+22
We add all the numbers together, and all the variables
8x^2-2x+16
Back to the equation:
-(8x^2-2x+16)
-8x^2+2x+(2x+3)^2+(2x-3)^2-16=0
We move all terms containing x to the left, all other terms to the right
-8x^2+2x+(2x+3)^2+(2x-3)^2=16
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