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

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Answer 1 · 2016-11-25T04:11:24

$n=+-(sqrt94 i)/3$

$9n^2-5=-99$

$9n^2=-94$

$n^2=-94/9$

$n=+-sqrt(-94/9)$

$n=+-(sqrt94 i)/3$

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