Сумма двух чисел равна 1,4 а сумма их квадратов равна 1. Найдите эти числа.

$x + y = 1.4 \\ {x}^{2} + {y}^{2} = 1 \\ \\ {(x + y)}^{2} = 1.96 \\ {x}^{2} + {y}^{2} = 1 \\ \\ {x}^{2} + 2xy + {y}^{2} = 1.96 \\ {x}^{2} + {y}^{2} = 1 \\ \\ 1 + 2xy = 1.96 \\ x + y = 1.4 \\ \\ 2xy = 0.96 \\ x + y = 1.4 \\ \\ xy = 0.48 \\ x + y = 1.4 \\ \\ y = 1.4 - x \\ x(1.4 - x) = 0.48 \\ \\ \\ y = 1.4 - x \\ 1.4x - {x}^{2} = 0.48 \\ \\ y = 1.4 - x \\ {x}^{2} - 1.4x + 0.48 = 0 \\ \\ d = {b}^{2} - 4ac = 1.96 - 4 \times 0.48 = 0.04 \\ x1 = \frac{1.4 + 0.2}{2} = \frac{1.6}{2} = 0.8 \\ x2 = \frac{1.4 - 0.2}{2} = \frac{1.2}{2} = 0.6 \\ \\ y1 = 1.4 - 0.8 = 0.6 \\ y2 = 1.4 - 0.6 = 0.8$
Ответ: (0,8; 0,6), (0,6; 0,8)