Question

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Answer 1

a)
${x}^{2} - 5x = 0 \\ x = \frac{ - ( - 5) + - \sqrt{ {( - 5)}^{2} - 4 \times 1 \times 0 } }{2 \times 1} \\ x = \frac{5 + - \sqrt{25} }{2} \\ x1 = \frac{5 + 5}{2} = \frac{10}{2} = 5 \\ x2 = \frac{5 - 5}{2} = 0$
b)
${x}^{2} + 5x + 6 = 0 \\ x = \frac{ - 5 + - \sqrt{ {( - 5)}^{2} - 4 \times 1 \times 6} }{2 \times 1} \\ x = \frac{ - 5 + - \sqrt{25 - 24} }{2} \\ x1 = \frac{ - 5 + 1}{2} = \frac{ - 4}{2} = - 2 \\ x2 = \frac{ - 5 - 1}{2} = \frac{ - 6}{2} = - 3$
c)
$3 {x}^{2} - 10x - 8 = 0 \\ x = \frac{ - ( - 10) + - \sqrt{ {( - 10)}^{2} - 4 \times 3 \times ( - 8) } }{2 \times 3} \\ x = \frac{10 + - \sqrt{100 + 96} }{6} \\ x1 = \frac{10 + 14}{6} = \frac{24}{6} = 4 \\ x2 = \frac{10 - 14}{6} = \frac{ - 4}{6} = \frac{ - 2}{3}$