2^2-2*y+y^2=16

Solution for 2^2-2*y+y^2=16 equation:

2^2-2y+y^2=16

We move all terms to the left:

2^2-2y+y^2-(16)=0

determiningTheFunctionDomain
y^2-2y-16+2^2=0

We add all the numbers together, and all the variables

y^2-2y-12=0

a = 1; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·1·(-12)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*1}=\frac{2-2\sqrt{13}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*1}=\frac{2+2\sqrt{13}}{2}
$