(1-2y)(1-2y)+y^2=1

Solution for (1-2y)(1-2y)+y^2=1 equation:

(1-2y)(1-2y)+y^2=1

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5y^2-4y=0

a = 5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·5·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*5}=\frac{0}{10}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*5}=\frac{8}{10}
=4/5
$