Question

пожалуйста, помогите​

Answer 1

$1)x {}^{2} + 5x + 7 = 0 \\ D = b {}^{2} - 4ac = 5 {}^{2} - 4 \times 1 \times 7 = 25 - 28 = - 3 \\ Пустое \: множество$

$3x {}^{2} + 8x + 24 = 0 \\ d = 64 - 288 = - 224 \\ Пустое \: множество$

$6x {}^{2} + 36x - 14 = 0 | \div 2 \\ 3x {}^{2} + 18x - 7 = 0 \\ d = (18) {}^{2} - 4 \times 3 \times ( - 7) = 408 = 2 \sqrt{102} \\ x1 = \frac{ - 18 + 2 \sqrt{102} }{6} = \frac{ - 9 + \sqrt{102} }{3} \\ x2 = \frac{ - 18 - 2 \sqrt{102} }{6} = \frac{ - 9 - \sqrt{102} }{3}$

$9x {}^{2} + 81x - 64 = 0 \\ d = 81 {}^{2} - 4 \times 9 \times ( - 64) = 6561 + 2304 = 8865 \\ x1 = \frac{ - 81 + 3 \sqrt{985} }{18} = \frac{ - 27 + \sqrt{985} }{6} \\ x2 = \frac{ - 18 - 3 \sqrt{985} }{18} = \frac{ - 27 - \sqrt{985} }{6}$

$36x {}^{2} + 75x + 4 = 0 \\ d = \sqrt{5049} = 3 \sqrt{561} \\ x1 = \frac{ - 75 + 3 \sqrt{561} }{72} = \frac{ - 25 + \sqrt{561} }{24} \\ x2 = \frac{ - 75 -3 \sqrt{561} }{72} = \frac{ - 25 - \sqrt{561} }{24}$

$16x {}^{2} + 33x + 28 = 0 \\ d = \sqrt{ - 703} \\ Пустое \: множество$