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

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}
$