(2x-1)2-(4x+1)(4x-1)=4x(1-2x)

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Solution for (2x-1)2-(4x+1)(4x-1)=4x(1-2x) equation:



(2x-1)2-(4x+1)(4x-1)=4x(1-2x)
We move all terms to the left:
(2x-1)2-(4x+1)(4x-1)-(4x(1-2x))=0
We add all the numbers together, and all the variables
(2x-1)2-(4x+1)(4x-1)-(4x(-2x+1))=0
We use the square of the difference formula
16x^2+(2x-1)2-(4x(-2x+1))+1=0
We multiply parentheses
16x^2+4x-(4x(-2x+1))-2+1=0
We calculate terms in parentheses: -(4x(-2x+1)), so:
4x(-2x+1)
We multiply parentheses
-8x^2+4x
Back to the equation:
-(-8x^2+4x)
We add all the numbers together, and all the variables
16x^2-(-8x^2+4x)+4x-1=0
We get rid of parentheses
16x^2+8x^2-4x+4x-1=0
We add all the numbers together, and all the variables
24x^2-1=0
a = 24; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·24·(-1)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*24}=\frac{0-4\sqrt{6}}{48} =-\frac{4\sqrt{6}}{48} =-\frac{\sqrt{6}}{12} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*24}=\frac{0+4\sqrt{6}}{48} =\frac{4\sqrt{6}}{48} =\frac{\sqrt{6}}{12} $

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