+958 Find the unique pair of real numbers (x,y) satisfying
(6x^2 - 18x + 17) + (3y^2 + 6y + 11) = 28
and x + y = 20.
Enter your answer as an ordered pair in the format $(x,y)$, where $x$ and $y$ are replaced by appropriate numbers.
+130477 6x^2 -18x + 17 + 3y^2 + 6y + 11 = 28
x + y = 20
y = 20 - x
So
6x^2 - 18x + 17 + 3(20 -x)^2 + 6(20 -x) + 11 = 28
6x^2 -18x + 3(x^2 - 40x + 400) + 120 -6x + 28 = 28
6x^2 - 18x + 3x^2 -120x + 1200 + 120 -6x = 0
9x^2 - 144x + 1320 = 0
No real solutions for this since the discriminant 144^2 - 36*1320 = -26784 is negative
