Odpowiedź:
[tex]c)2 {x}^{2} + 5x - 12 = 0 \\a = 2 \\ b = 5 \\ c = - 12 \\ ∆ = {b}^{2} - 4ac = {5}^{2} - 4 \times 2 \times ( - 12) = 25 + 96 = 121 \\ \sqrt{∆} = \sqrt{121} = 11 \\ x_{1} = \frac{ - b - \sqrt{∆} }{2a} = \frac{ - 5 - 11}{2 \times 2} = \frac{ - 16}{4} = - 4 \\ x_{2} = \frac{ - b + \sqrt{∆} }{2a} = \frac{ - 5 + 11}{4} = \frac{6}{4} = \frac{3}{2} [/tex]
[tex]d)3 {x}^{2} + 7x - 20 = 0 \\ a = 3 \\ b = 7 \\ c = - 20 \\ ∆ = {b}^{2} - 4ac = {7}^{2} - 4 \times 3 \times ( - 20) = 49 + 240 = 289 \\ \sqrt{∆} = \sqrt{289} = 17 \\ x_{1} = \frac{ - b - \sqrt{∆} }{2 a} = \frac{ - 7 - 17}{6} = \frac{ - 24}{6} = - 4 \\ x_{2} = \frac{ - b + \sqrt{∆} }{2a} = \frac{ - 7 + 17}{6} = \frac{10}{6} = \frac{5}{3} [/tex]
[tex]c)16 {x}^{2} - 56x + 49 = 0 \\ a = 16 \\ b = - 56 \\ c = 49 \\ ∆ = {b}^{2} - 4ac = ( - 56 {)}^{2} - 4 \times 16 \times 49 = 3136 - 3136 = 0 \\ x_{0} = \frac{ - b}{2a} = \frac{56}{32} = 1 \frac{3}{4} [/tex]
[tex]d) - {x}^{2} - 2x + 2 = 0 \\ a = - 1 \\ b = - 2 \\ c = 2 \\ ∆ = {b}^{2} - 4ac = ( - 2 {)}^{2} - 4 \times2 \times ( - 1) = 4 + 8 = 12 \\ \sqrt{∆} = \sqrt{12} = 2 \sqrt{3} \\ x_{1} = \frac{ - b - \sqrt{∆} }{2a} = \frac{2 - 2 \sqrt{3} }{ - 2} = \frac{ - 2( - 1 + \sqrt{3}) }{ - 2} = - 1 + \sqrt{3} \\ x_{2} = \frac{ - b + \sqrt{∆} }{2a} = \frac{2 + 2 \sqrt{3} }{ - 2} = \frac{2(1 + \sqrt{3}) }{ - 2} = - 1 - \sqrt{3} [/tex]
