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

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}
$