4y(2y+1)=2y+1

Solution for 4y(2y+1)=2y+1 equation:

4y(2y+1)=2y+1

We move all terms to the left:

4y(2y+1)-(2y+1)=0

We multiply parentheses

8y^2+4y-(2y+1)=0

We get rid of parentheses

8y^2+4y-2y-1=0

We add all the numbers together, and all the variables

8y^2+2y-1=0

a = 8; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·8·(-1)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*8}=\frac{-8}{16}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*8}=\frac{4}{16}
=1/4
$