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

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Solution for (4x+5)(4x-5)-(2x-3)^2-3(x+5)^2-7=0 equation: 



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

We use the square of the difference formula

16x^2-(2x-3)^2-3(x+5)^2-25-7=0

We add all the numbers together, and all the variables

16x^2-(2x-3)^2-3(x+5)^2-32=0

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

16x^2-(2x-3)^2-3(x+5)^2=32

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