Question

$3x {}^{2} - 2 \sqrt{6x } + 2 = 0$
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Answer 1

Answer:

$3x - 2 \sqrt{6x} + 2$

${ (\sqrt{3}x) }^{2} - 2. \sqrt{3x}. \sqrt{2} + {( \sqrt{2} )}^{2}$

$by \: the \: property \: {(a - b)}^{2}$

${( \sqrt{3x} - \sqrt{2})}^{2}$

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Answer 2

see the above answer that is correct