How do you solve the quadratic $2n^2-5=-35$ using any method?

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Answer 1 · 2016-12-31T05:57:07

$n=+-sqrt(15)i$

Solve $2n^2-5=-35$ .

Add $5$ to both sides.

$2n^2=-30$

Divide both sides by $2$ .

$n^2=-30/2$

$n^2=-15$

Take the square root of both sides.

$n=+-sqrt(-15)$

$n=+-sqrt(15)i$

How do you solve the quadratic 2n^2-5=-352n^2-5=-35 using any method?