Question

How do you solve the quadratic with complex numbers given 27c2−12c−314=0?

Answer 1

$27c {}^{2} - 12c - 314 = 0 \\ = > (\sqrt{27} c) {}^{2} - 2 \times \sqrt{27} c \times \frac{6}{ \sqrt{27} } + (\frac{6}{ \sqrt{27} }) {}^{2} - 314 = (\frac{6}{ \sqrt{27} }) {}^{2} \\ = > ( \sqrt{27} c - \frac{6}{ \sqrt{27} } ) {}^{2} - 314 = \frac{36}{27} \\ = > ( \sqrt{27} c - \frac{6}{ \sqrt{27} } ) {}^{2} = \frac{36}{27} + 314 \\ = > ( \sqrt{27} c - \frac{6}{ \sqrt{27} } ) {}^{2} = \frac{36 + 8478}{27} \\ = > ( \sqrt{27} c - \frac{6}{ \sqrt{27} } ) {}^{2}= \frac{8514}{27} \\ = > ( \sqrt{27} c - \frac{6}{ \sqrt{27} } ) {}^{2} = \frac{946}{3} \\ = > \sqrt{27} c - \frac{6}{3 \sqrt{3} } = ( \frac{ + }{ - } ) \sqrt{ \frac{946}{3} } \\ = > \sqrt{27} c = \frac{2}{ \sqrt{3} } ( \frac{ + }{ - } )\sqrt{ \frac{946 }{3} } \\ = > \sqrt{27} c = \frac{2}{ \sqrt{3} } (\frac{ + }{ - } ) \frac{ \sqrt{946} }{ \sqrt{3} } \\ = > 3 \sqrt{3} c = \frac{2( + - ) \sqrt{946} }{ \sqrt{3} } \\ = > c = \frac{2( + - ) \sqrt{946} }{ \sqrt{3} \times 3 \sqrt{3} } \\ = > c = \frac{2( + - ) \sqrt{946} }{9} \\ = > c = \frac{2 + \sqrt{946} }{9} \: or \: c = \frac{2 - \sqrt{946} }{9} \: ans...$

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