How do you solve the quadratic $9 n^{2} - 5 = - 99$ $9n^2-5=-99$ using any method?

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Answer 1 · 2016-11-25T04:11:24

$n = \pm \frac{\sqrt{94} i}{3}$ $n=+-(sqrt94 i)/3$

$9 n^{2} - 5 = - 99$ $9n^2-5=-99$

$9 n^{2} = - 94$ $9n^2=-94$

$n^{2} = - \frac{94}{9}$ $n^2=-94/9$

$n = \pm \sqrt{- \frac{94}{9}}$ $n=+-sqrt(-94/9)$

$n = \pm \frac{\sqrt{94} i}{3}$ $n=+-(sqrt94 i)/3$

How do you solve the quadratic 9n2−5=−999n^2-5=-99 using any method?