Решите систему уравнений х квадрат+у квадрат=26 ху=-5

$x^2 + y^2 = 26\\ xy = -5\\\\ (x + y)^2 - 2xy = 26\\ (x + y)^2 +10 = 26\\\\ (x+y)^2 = 16\\\\ x+y = 4, x+y = -4\\\\ 1) y(4-y) = -5\\ 4y - y^2 + 5 = 0\\ y^2 - 4y - 5 = 0\\ y_1 + y_2 = 4\\ y_1 * y_2 = -5\\ y_1 = -1, y_2 = 5\\ x = 5, x = -1\\\\ 2) y(-4-y) = -5\\ 4y + y^2 = 5\\ y^2 + 4y - 5 = 0\\ y_1 + y_2 = -4\\ y_1 * y_2 = -5\\ y_1 = 1, y_2 = -5\\ x = -5, x = 1\\\\ (5 , -1), (-1, 5), (1, -5), (-5, 1)$