(4x+1)(8x-1)=16x+3

Solution for (4x+1)(8x-1)=16x+3 equation:

(4x+1)(8x-1)=16x+3

We move all terms to the left:

(4x+1)(8x-1)-(16x+3)=0

We get rid of parentheses

(4x+1)(8x-1)-16x-3=0

We multiply parentheses ..

(+32x^2-4x+8x-1)-16x-3=0

We get rid of parentheses

32x^2-4x+8x-16x-1-3=0

We add all the numbers together, and all the variables

32x^2-12x-4=0

a = 32; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·32·(-4)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{41}}{2*32}=\frac{12-4\sqrt{41}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{41}}{2*32}=\frac{12+4\sqrt{41}}{64}
$