(4x)(x+1)=1

Solution for (4x)(x+1)=1 equation:

(4x)(x+1)=1

We move all terms to the left:

(4x)(x+1)-(1)=0

We multiply parentheses

4x^2+4x-1=0

a = 4; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·4·(-1)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{2}}{2*4}=\frac{-4-4\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{2}}{2*4}=\frac{-4+4\sqrt{2}}{8}
$