(12-y)*y=32 Решите плиз
y = 8 и y = 4
(12-у) × у = 32 12у - у^2 = 32 -у^2 + 12 - 32 = 0 | × (-1) у^2 - 12 + 32 = 0 D = b^2 - 4ac =12^2 - 4 × 1 × 32 = 144 - 128 = 16 y1 = (12 - 4) / 2 = 4 y2 = (12 + 4) / 2 = 8 Ответ: у = 4; у = 8.