(2x-1)(4x+1)=(8x+1)=(8x-3)(x-2)

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Solution for (2x-1)(4x+1)=(8x+1)=(8x-3)(x-2) equation:



(2x-1)(4x+1)=(8x+1)=(8x-3)(x-2)
We move all terms to the left:
(2x-1)(4x+1)-((8x+1))=0
We multiply parentheses ..
(+8x^2+2x-4x-1)-((8x+1))=0
We calculate terms in parentheses: -((8x+1)), so:
(8x+1)
We get rid of parentheses
8x+1
Back to the equation:
-(8x+1)
We get rid of parentheses
8x^2+2x-4x-8x-1-1=0
We add all the numbers together, and all the variables
8x^2-10x-2=0
a = 8; b = -10; c = -2;
Δ = b2-4ac
Δ = -102-4·8·(-2)
Δ = 164
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:
$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{41}}{2*8}=\frac{10-2\sqrt{41}}{16} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{41}}{2*8}=\frac{10+2\sqrt{41}}{16} $

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