Question

Solve the following quadratic equation by quadratic formula. 2x square +5√3x +6 =0​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

$\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{Roots \ are \ -2\sqrt3 \ and \ \frac{-\sqrt3}{2}}$

$\orange{Given:}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{2x^{2}+5\sqrt3x+6=0}}$

$\pink{To \ find:}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{\implies{2x^{2}+5\sqrt3x+6=0}}$

$\sf{Here,}$

$\sf{a=2}$

$\sf{b=5\sqrt3}$

$\sf{c=6}$

$\sf{By \ formula \ method}$

$\sf{b^{2}-4ac=(5\sqrt3)^{2}-4(2)(6)}$

$\sf{=75-48}$

$\sf{=27}$

$\sf{x=\frac{-b-\sqrt(b^{2}-4ac)}{2a} \ or \ \frac{-b+\sqrt(b^{2}-4ac)}{2a}}$

$\sf{x=\frac{-5\sqrt3-\sqrt27}{2×2} \ or \ \frac{-5\sqrt3+\sqrt27}{2×2}}$

$\sf{x=\frac{-5\sqrt3-3\sqrt3}{4} \ or \ \frac{-5\sqrt3+3\sqrt3}{4}}$

$\sf{x=\frac{-8\sqrt3}{4} \ or \ \frac{-2\sqrt3}{4}}$

$\sf{x=-2\sqrt3 \ or \ \frac{-\sqrt3}{2}}$

$\sf\purple{\tt{\therefore{Roots \ are \ -2\sqrt3 \ and \ \frac{-\sqrt3}{2}}}}$