(1-y)(1-y)+y^2=22

Solution for (1-y)(1-y)+y^2=22 equation:

(1-y)(1-y)+y^2=22

We move all terms to the left:

(1-y)(1-y)+y^2-(22)=0

We add all the numbers together, and all the variables

y^2+(-1y+1)(-1y+1)-22=0

We multiply parentheses ..

y^2+(+y^2-1y-1y+1)-22=0

We get rid of parentheses

y^2+y^2-1y-1y+1-22=0

We add all the numbers together, and all the variables

2y^2-2y-21=0

a = 2; b = -2; c = -21;
Δ = b2-4ac
Δ = -22-4·2·(-21)
Δ = 172
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{43}}{2*2}=\frac{2-2\sqrt{43}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{43}}{2*2}=\frac{2+2\sqrt{43}}{4}
$