(16x^2-4x-1)/(2x+1)=0

Solution for (16x^2-4x-1)/(2x+1)=0 equation:

(16x^2-4x-1)/(2x+1)=0


Domain of the equation: (2x+1)!=0

We move all terms containing x to the left, all other terms to the right

2x!=-1

x!=-1/2

x!=-1/2

x∈R


We multiply all the terms by the denominator


(16x^2-4x-1)=0

We get rid of parentheses

16x^2-4x-1=0

a = 16; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·16·(-1)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{5}}{2*16}=\frac{4-4\sqrt{5}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{5}}{2*16}=\frac{4+4\sqrt{5}}{32}
$