Question

6
$6x {}^{2} + x - 1 = 0$
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Answer 1

$6 {x}^{2} + x - 1 = 0 \\ delta \: \sqrt{ {1}^{2} - 4 \times 6 \times ( - 1) } \\ delta = \sqrt{25} \\ delta = 5 \\ x = \frac{ - 1 + - 5}{2 \times 6} \\ x = - \frac{ - 1 + - 5}{12} \\ x1 = \frac{ - 4}{12} = - \frac{1}{3} \\ x2 = - \frac{ 6}{12} = - \frac{1}{2}$