How to find point C(x,y)

C(x,y) is part of the equilateral triangle ABC: A(2,-5), B(3,-4).

the distances are 2^(1/2), but i can't find x and y individually;
help pls

(x - 2)^2 + (y + 5)^2 = 2
(x - 3)^2 + (y + 4)^2 = 2

x^2 - 4x + 4 + y^2 + 10y + 25 = 2
x^2 - 6x + 9 + y^2 + 8y + 16 = 2

2x - 5 + 2y + 9 = 2
2x + 2y = -4
x + y = -2
y = -2 - x

(x - 2)^2 + (y + 5)^2 = 2
(x - 2)^2 + (-2 - x + 5)^2 = 2
(x - 2)^2 + (-x + 3)^2 = 2
x^2 - 4x + 4 + x^2 - 6x + 9 = 2
2x^2 - 10x + 11 = 0
x = (5 ± √3)/2
y = (-9 ∓ √3)/2

C(x,y) = ([5 ± √3]/2, [-9 ∓ √3]/2)
Note that there are two possibilities.