Question

3x^2+2x=0 7x^2+6x-1=0 2x^2-^18=0 2x^2-x+11=0 хелп,я аут,

Answer 1

$3 {x}^{2} + 2x = 0 \\ x(3x + 2) = 0 \\ x = 0 \\ x = \frac{ - 2}{3} \\ \\ 7 {x}^{2} + 6x - 1 = 0 \\ d1 = {k}^{2} - ac = 9 + 7 = 16 \\ x1 = \frac{ - k + \sqrt{d1} }{a} = \frac{ - 3 + 4}{7} = \frac{1}{7} \\ x2 = \frac{ - k - \sqrt{d1} }{a} = - 1 \\ \\ 2 {x}^{2} - 18 = 0 \\ {x}^{2} = \frac{18}{2} \\ {x}^{2} = 9 \\ x1 = 3 \\ x2 = - 3 \\ \\ 2 {x}^{2} - x + 11 = 0 \\ d = {b}^{2} - 4ac = 1 - 4 \times 2 \times 11 < 0$
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