603. Найдите корни квадратного трёхчлена: a) 10x² + 5x – 5; в) x² – 2x – 4; б) -2x² + 12x – 18; г) 12x² – 12.

а) 10x² + 5x - 5 = 0 D = 5² - 4 * 10 * (-5) = 25 + 200 = 225 x₁ = (-5 + √225) / (2 * 10) = (-5 + 15) / 20 = 10 / 20 = 1 / 2 = 0,5 x₂ = (-5 - √225) / (2 * 10) = (-5 - 15) / 20 = -20 / 20 = -1 б) -2x² + 12x - 18 = 0 D = 12² - 4 * (-2) * (-18) = 144 - 144 = 0 x = -12 / (2 * (-2)) = -12 / (-4) = 3 в) x² - 2x - 4 = 0 D = (-2)² - 4 * 1 * (-4) = 4 + 16 = 20 x₁ = (2 + √20) / (2 * 1) = (2 + 2√5) / 2 = 1 + √5 x₂ = (2 - √20) / (2 * 1) = (2 - 2√5) / 2 = 1 - √5 г) 12x² - 12 = 0 12x² = 12 x² = 12 / 12 = 1 x = ±1 Ответ: a) 0,5, -1; б) 3; в) 1 + √5, 1 - √5; г) 1, -1