Question

Solve using Quadratic formula​

Answer 1

Your Answer in the Attachment^

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Answer 2

$2x {}^{2} + 5 \sqrt{} 3x + 6 =0 \\ a = 2 \: \: b = 5 \sqrt{} 3 \: \: c = 6 \\ d = b {}^{2} - 4ac \\ = (5 \sqrt{3} ) {}^{2} - 4(2)(6) \\ = 75 - 48 = 27 > 0 \\ since \: d > 0 \: there \: for \: the \: given \: quadratic \: equation \: has \: real \: roots \: \\ now \: \\ 2x {}^{2} + 5 \sqrt{} 3 + 6 = 0 \\ = 2x {}^{2} + 4 \sqrt{} 3 + \sqrt{} 3x + 6 = 0 \\ = 2x(x + 2 \sqrt{3) + \sqrt{} } 3(x + 2 \sqrt{} 3) = 0 \\ = (x + 2 \sqrt{} 3) \: (2x + 2 \sqrt{} 3) = 0 \\ x = - 2 \sqrt{} 3 \: or \: - \sqrt{} 3 \div 2$

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