Question

solve this problem plzzz

Answer 1

$2x { }^{2} - 2 \sqrt{6x } + 3 = 0$
this is in the form of Q.E so by the formula
$b {}^{2} - 4ac$
$a = 2. \: b = 2 \sqrt{6x}. \: c = 3.$
now
$< 2 \sqrt{6 {}^{2} } > {}^{2} - 4 \times 2 \times 3$
$2 \times 6 - 24$
$12 - 24 = 12$
I hope it help u