Help solve and explain

Solve the system:

x^2 + y^2 = 23,

x+y = 1

x + y = 1
x² + 2xy + y² = 1
23 + 2xy = 1
xy = -11

y = 1 - x

x(1-x) = -11
x - x² = -11
x² - x = 11
(x - 1/2)² = 45/4
x - 1/2 = ±(45) / 2
x = 1/2 ± 3√(5) / 2
x = (1 ± 3 √(5)) / 2

x + y = 1
y = 1 - (1 ± 3√(5)) / 2

x + y = 1 → y = 1 - x

... x^2 + y^2 = 23
or x^2 + (1 - x)^2 = 23
or x^2 + (x^2 - 2x +1) = 23
or 2x^2 - 2x - 22 = 0
or x^2 - x - 11 = 0

The Quadratic Formula:
x = [ -b ± √ ( (b)² - 4(a)(c) ) ] / 2(a)
x = [ -(-1) ± √ ( (-1)² - 4(1)(-11) ) ] / 2(1)
x = [ 1 ± √ ( 1 + 44 ) ] / 2
x = ½ ( 1 ± √45 )
x = ½ ( 1 ± 3√5 )


x = ½ ( 1 - 3√5 ) ... → y = 1 - ½ ( 1 - 3√5 ) = ½ ( 1 + 3√5 )
x = ½ ( 1 + 3√5 ) .. → y = 1 - ½ ( 1 + 3√5 ) = ½ ( 1 - 3√5 )

(x,y) = { ( ½ ( 1-3√5 ), ½ ( 1+3√5 ) ), ( ½ ( 1+3√5 ), ½ ( 1-3√5 ) ) }