Question

n^2 + n + 52. = 0

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Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

Given Equation,

$\large{\sf{n {}^{2} + n + 52 = 0}}$

On comparing with ax² + bx + c = 0,

a = 1 ,b = 1 and c = 52

Discriminant,d:

$\sf{d = b {}^{2} - 4ac} \\ \\ \rightarrow \: \sf{d = (1) {}^{2} - 4(1)(52)} \\ \\ \rightarrow \underline{\boxed{\sf{d = - \: 207}}} \:$

Applying the Quadratic Formula,

$\sf{n = \frac{ - b \pm \sqrt{d} }{2a} } \\ \\ \implies \: \sf{n = \frac{ - 1 \: \pm \: \sqrt{ - 207} }{2} } \\ \\ \implies \ \: \sf{n = \frac{ - 1 \: \pm \: 3 \sqrt{23}i }{2} } \\ \\ \implies \ \: \boxed{\sf{n = \frac{ - 1 \: - 3\sqrt{23}i }{2} \: \: or \: \: \frac{ - 1 \: + \: 3 \sqrt{23}i }{2} }}$

Answer 2

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