Question

find the roots of
$3x ^{2} - 2 \sqrt{6x} + 2 = 0$

Answer 1

$3 {x}^{2} - 2 \sqrt{6} x + 2 = 0 \\ \\ 3 {x}^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0 \\ \\ \sqrt{3} x( \sqrt{3} x - \sqrt{2} ) - \sqrt{2} ( \sqrt{3} x - \sqrt{2} ) = 0 \\ \\ ( \sqrt{3x} - \sqrt{2} )( \sqrt{3} x - \sqrt{2} ) = 0$
$x = \frac{ \sqrt{2} }{ \sqrt{3} }$