Question

solve the quadratic equations with factorization method
$6 \sqrt{3} {x}^{2} + 7x = \sqrt{3}$
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Answer 1

$6 \sqrt{3} {x}^{2} + 7x - \sqrt{3} = 0 \\ = 6 \sqrt{3} {x}^{2} + 9x - 2x - \sqrt{3} = 0 \\ = 3 \sqrt{3}x (2x + 3) - 1(2x - 3) = 0 \\ = (2x - 3)(3 \sqrt{3}x - 1) = 0 \\ either \: 2x - 3 = 0 \\ = > 2x = 3 \\ = > x = \frac{3}{2} \\ or \: 3 \sqrt{3}x - 1 = 0 \\ = > 3 \sqrt{3} x = 1 \\ = > x = \frac{1}{3 \sqrt{3} } (ans)$

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Answer 2

Answer:

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