+2653 Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 32 + 2xy$.
For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).
+129850 x + y = 10
y = 10 -x
x^2 + (10 -x)^2 = 32 + 2x(10-x)
x^2 + x^2 -20x + 100 = 32 + 20x -2x^2
4x^2 -40x + 68 = 0
x^2 -10x + 17 = 0
x^2 -10x = -17
x^2 -10x + 25 = -17 + 25
(x -5)^2 = 8
x -5 = +/- sqrt (8) = +/- 2sqrt 2
x = 5 + 2sqrt2 or x = 5 - 2sqrt 2
y = 10 -x = 5 -2sqrt 2 or 5 + 2sqrt 2
(x,y) = (5 + 2sqrt 2 , 5-2sqrt 2) or ( 5- 2sqrt 2, 5 + 2sqrt 2)
