Let (x, y) be an ordered pair of real numbers that satisfies the equation \(x^{2}+y^{2}=14x+48y \). What is the minimum value of x?
Nvm, I solved it
2(7x+24y)=x^2+y^2
x^2-14x=48y-y^2
x(x-14)=y(48-y)
(x,y)=(0,0) , (14,48) , (0,48),(14,0) etc..
+296 
