5. 4 С помощью формулы корней квадратного уравнения решить уравнение: 1) x²+4x-21 = 0; 3) 2x²+x-21 = 0; 5) 6x²+19x-7=0; 2) x²-2x-24 = 0; 4) 5x²+14x-24 = 0; 6) 10x²-7x-6=0.

1) x²+4x-21 = 0

• a = 1, b = 4, c = -21
• \( D = 4^2 - 4 \cdot 1 \cdot (-21) = 16 + 84 = 100 \)
• \( x_1 = \frac{-4 + \sqrt{100}}{2 \cdot 1} = \frac{-4 + 10}{2} = \frac{6}{2} = 3 \)
• \( x_2 = \frac{-4 - \sqrt{100}}{2 \cdot 1} = \frac{-4 - 10}{2} = \frac{-14}{2} = -7 \)

Ответ: x₁ = 3, x₂ = -7

2) x²-2x-24 = 0

• a = 1, b = -2, c = -24
• \( D = (-2)^2 - 4 \cdot 1 \cdot (-24) = 4 + 96 = 100 \)
• \( x_1 = \frac{2 + \sqrt{100}}{2 \cdot 1} = \frac{2 + 10}{2} = \frac{12}{2} = 6 \)
• \( x_2 = \frac{2 - \sqrt{100}}{2 \cdot 1} = \frac{2 - 10}{2} = \frac{-8}{2} = -4 \)

Ответ: x₁ = 6, x₂ = -4

3) 2x²+x-21 = 0

• a = 2, b = 1, c = -21
• \( D = 1^2 - 4 \cdot 2 \cdot (-21) = 1 + 168 = 169 \)
• \( x_1 = \frac{-1 + \sqrt{169}}{2 \cdot 2} = \frac{-1 + 13}{4} = \frac{12}{4} = 3 \)
• \( x_2 = \frac{-1 - \sqrt{169}}{2 \cdot 2} = \frac{-1 - 13}{4} = \frac{-14}{4} = -3.5 \)

Ответ: x₁ = 3, x₂ = -3.5

4) 5x²+14x-24 = 0

• a = 5, b = 14, c = -24
• \( D = 14^2 - 4 \cdot 5 \cdot (-24) = 196 + 480 = 676 \)
• \( x_1 = \frac{-14 + \sqrt{676}}{2 \cdot 5} = \frac{-14 + 26}{10} = \frac{12}{10} = 1.2 \)
• \( x_2 = \frac{-14 - \sqrt{676}}{2 \cdot 5} = \frac{-14 - 26}{10} = \frac{-40}{10} = -4 \)

Ответ: x₁ = 1.2, x₂ = -4

5) 6x²+19x-7=0

• a = 6, b = 19, c = -7
• \( D = 19^2 - 4 \cdot 6 \cdot (-7) = 361 + 168 = 529 \)
• \( x_1 = \frac{-19 + \sqrt{529}}{2 \cdot 6} = \frac{-19 + 23}{12} = \frac{4}{12} = \frac{1}{3} \)
• \( x_2 = \frac{-19 - \sqrt{529}}{2 \cdot 6} = \frac{-19 - 23}{12} = \frac{-42}{12} = -3.5 \)

Ответ: x₁ = 1/3, x₂ = -3.5

6) 10x²-7x-6=0

• a = 10, b = -7, c = -6
• \( D = (-7)^2 - 4 \cdot 10 \cdot (-6) = 49 + 240 = 289 \)
• \( x_1 = \frac{7 + \sqrt{289}}{2 \cdot 10} = \frac{7 + 17}{20} = \frac{24}{20} = 1.2 \)
• \( x_2 = \frac{7 - \sqrt{289}}{2 \cdot 10} = \frac{7 - 17}{20} = \frac{-10}{20} = -0.5 \)

Ответ: x₁ = 1.2, x₂ = -0.5