Hello, Vince604!
Solve:
. . x^{2} + y^{2} = 20 . [1]
. . . . y = x^{2} . . . [2]
Substitute [2] into [1]: .y + y^{2} = 20 . → . y^{2} + y - 20 = 0
. . . (y - 4)(y + 5) .= .0 . → . y = 4, -5
Substitute into [2]:
. . 4 = x^{2} . → . x = +2
. .-5 = x^{2} . No real roots
Therefore: .(x, y) .= .(+2, 4)