Question

n^2 + n 52 = 0

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Answer 1

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${n}^{2} + n + 52 = 0. \\ \\ a = 1 \: \: \: \: \: \: \: \: \: \: \: b = 1 \: \: \: \: \: \: \: \: \: \: c = 52. \\ \\ d = {b}^{2} - 4ac. \\ \\ {(1)}^{2} - 4 \times 1 \times 52. \\ \\ = 1 - 208. \\ \\ - 207. \\ \\ \\ \\ n = \frac{ - b ± \sqrt{ {b}^{2} - 4ac } }{2a} . \\ \\ n = \frac{ - 1± \sqrt{ - 207} }{2} .$

Now,

$let \: \: \: \: \: \: \: \: \: \: \: \: \sqrt{ - 1} = i.$

Putting the value. :-

$n= \frac{ - 1±\sqrt{207i} }{2} . \\ \\ n= \frac{ - 1± 3 \sqrt{23i} }{2} .$

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Answer 2

Answer:

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Step-by-step explanation:

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