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(3x-7)^2-(5x-2)(5x+2)+32=(4x-1)(4x-1)
We move all terms to the left:
(3x-7)^2-(5x-2)(5x+2)+32-((4x-1)(4x-1))=0
We use the square of the difference formula
25x^2+(3x-7)^2-((4x-1)(4x-1))+4+32=0
We multiply parentheses ..
25x^2-((+16x^2-4x-4x+1))+(3x-7)^2+4+32=0
We calculate terms in parentheses: -((+16x^2-4x-4x+1)), so:We add all the numbers together, and all the variables
(+16x^2-4x-4x+1)
We get rid of parentheses
16x^2-4x-4x+1
We add all the numbers together, and all the variables
16x^2-8x+1
Back to the equation:
-(16x^2-8x+1)
25x^2-(16x^2-8x+1)+(3x-7)^2+36=0
We get rid of parentheses
25x^2-16x^2+8x+(3x-7)^2-1+36=0
We add all the numbers together, and all the variables
9x^2+8x+(3x-7)^2+35=0
We move all terms containing x to the left, all other terms to the right
9x^2+8x+(3x-7)^2=-35
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