Question

plz solve it...... ​

Answer 1

$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

$\sf\implies x^2 - \sqrt 3 x - 6 = 0$

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compare the eq with $\sf{\underline{\bold{ax^2 + bx + c = 0 }}}$

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☯ a = 1

☯ b = -√3

☯ c = -6

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now :-

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$\sf{\underline{\boxed{\pink{\large{\mathfrak{x = \dfrac{ - b \pm \sqrt D }{2a }}}}}}}$

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$\sf{\underline{\boxed{\pink{\large{\mathfrak{ D = b^2 - 4ac }}}}}}$

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finding value of D.

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$\sf\implies D = b^2 - 4ac$

$\sf\implies D = (-√3)^2 - 4 \times 1 \times -6$

$\sf\implies D = 3 + 24$

$\sf\implies D = 27$

$\sf{\underline{\boxed{\blue{\large{\bold{ D = 27}}}}}}$

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putting values in the eq.

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$\sf\implies x = \dfrac{ -( -\sqrt 3) \pm\sqrt {27} }{2\times 1 }$

$\sf\implies x = \dfrac{ \sqrt 3 \pm 3\sqrt 3}{2}$

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✒$\sf x = \dfrac{ \sqrt 3 + 3\sqrt 3 }{ 2 }$

$\implies x = \dfrac {4\sqrt 3}{2}$

$\implies x = 2\sqrt 3$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = 2\sqrt 3 }}}}}}$

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✒$\sf x = \dfrac{ \sqrt 3 - 3\sqrt 3 }{ 2 }$

$\implies x = \dfrac {-2\sqrt 3}{2}$

$\implies x = - \sqrt 3$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = - \sqrt 3 }}}}}}$

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$\sf{\underline{\boxed{\purple{\large{\bold{ x = 2\sqrt 3 \: or \:- \sqrt 3 }}}}}}$

Answer 2

Step-by-step explanation:

x^2-√3x-6 =0

x^2-2√3x+√3x-6=0

x(x-2√3)+√3(x-2√3)=0

(x-2√3)(x+√3)=0

x=2√3, (-√3)

please mark brainliest