x² + 8 x-33 = 0 x² + 12 x - 64 = 0 x²-11 x + 30 = 0 x² + 14 x + 24 = 0 x2 - x - 20 = 0 x²+x-30 = 0 x²-x-12 = 0 x²-7 x + 12 = 0 x² - 2,4 x - 13 = 0 x² - 5,6 x + 6,4 = 0

Решим уравнения по порядку, используя дискриминант. 1) \(x^2 + 8x - 33 = 0\) \[D = b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot (-33) = 64 + 132 = 196\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{196}}{2 \cdot 1} = \frac{-8 + 14}{2} = \frac{6}{2} = 3\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{196}}{2 \cdot 1} = \frac{-8 - 14}{2} = \frac{-22}{2} = -11\] 2) \(x^2 + 12x - 64 = 0\) \[D = b^2 - 4ac = 12^2 - 4 \cdot 1 \cdot (-64) = 144 + 256 = 400\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-12 + \sqrt{400}}{2 \cdot 1} = \frac{-12 + 20}{2} = \frac{8}{2} = 4\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-12 - \sqrt{400}}{2 \cdot 1} = \frac{-12 - 20}{2} = \frac{-32}{2} = -16\] 3) \(x^2 - 11x + 30 = 0\) \[D = b^2 - 4ac = (-11)^2 - 4 \cdot 1 \cdot 30 = 121 - 120 = 1\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{11 + \sqrt{1}}{2 \cdot 1} = \frac{11 + 1}{2} = \frac{12}{2} = 6\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{11 - \sqrt{1}}{2 \cdot 1} = \frac{11 - 1}{2} = \frac{10}{2} = 5\] 4) \(x^2 + 14x + 24 = 0\) \[D = b^2 - 4ac = 14^2 - 4 \cdot 1 \cdot 24 = 196 - 96 = 100\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-14 + \sqrt{100}}{2 \cdot 1} = \frac{-14 + 10}{2} = \frac{-4}{2} = -2\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-14 - \sqrt{100}}{2 \cdot 1} = \frac{-14 - 10}{2} = \frac{-24}{2} = -12\] 5) \(x^2 - x - 20 = 0\) \[D = b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{81}}{2 \cdot 1} = \frac{1 + 9}{2} = \frac{10}{2} = 5\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{81}}{2 \cdot 1} = \frac{1 - 9}{2} = \frac{-8}{2} = -4\] 6) \(x^2 + x - 30 = 0\) \[D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-30) = 1 + 120 = 121\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{121}}{2 \cdot 1} = \frac{-1 + 11}{2} = \frac{10}{2} = 5\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{121}}{2 \cdot 1} = \frac{-1 - 11}{2} = \frac{-12}{2} = -6\] 7) \(x^2 - x - 12 = 0\) \[D = b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-12) = 1 + 48 = 49\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{49}}{2 \cdot 1} = \frac{1 + 7}{2} = \frac{8}{2} = 4\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{49}}{2 \cdot 1} = \frac{1 - 7}{2} = \frac{-6}{2} = -3\] 8) \(x^2 - 7x + 12 = 0\) \[D = b^2 - 4ac = (-7)^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{1}}{2 \cdot 1} = \frac{7 + 1}{2} = \frac{8}{2} = 4\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{1}}{2 \cdot 1} = \frac{7 - 1}{2} = \frac{6}{2} = 3\] 9) \(x^2 - 2.4x - 13 = 0\) \[D = b^2 - 4ac = (-2.4)^2 - 4 \cdot 1 \cdot (-13) = 5.76 + 52 = 57.76\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2.4 + \sqrt{57.76}}{2 \cdot 1} = \frac{2.4 + 7.6}{2} = \frac{10}{2} = 5\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2.4 - \sqrt{57.76}}{2 \cdot 1} = \frac{2.4 - 7.6}{2} = \frac{-5.2}{2} = -2.6\] 10) \(x^2 - 5.6x + 6.4 = 0\) \[D = b^2 - 4ac = (-5.6)^2 - 4 \cdot 1 \cdot 6.4 = 31.36 - 25.6 = 5.76\] \[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5.6 + \sqrt{5.76}}{2 \cdot 1} = \frac{5.6 + 2.4}{2} = \frac{8}{2} = 4\] \[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5.6 - \sqrt{5.76}}{2 \cdot 1} = \frac{5.6 - 2.4}{2} = \frac{3.2}{2} = 1.6\]

Ответ:

1) x = 3, x = -11

2) x = 4, x = -16

3) x = 6, x = 5

4) x = -2, x = -12

5) x = 5, x = -4

6) x = 5, x = -6

7) x = 4, x = -3

8) x = 4, x = 3

9) x = 5, x = -2.6

10) x = 4, x = 1.6

Все получилось просто отлично! Ты хорошо умеешь решать квадратные уравнения, продолжай в том же духе!