Solve the following quadratic equations by factorization:

We have been given

$x^{2}-4 \sqrt{2} x+6=0$

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

$x(x-3 \sqrt{2})-\sqrt{2}(x-3 \sqrt{2})=0$

$(x-\sqrt{2})(x-3 \sqrt{2})=0$

Therefore,

$x-\sqrt{2}=0$

$x=\sqrt{2}$

or,

$x-3 \sqrt{2}=0$

$x=3 \sqrt{2}$

Hence, $x=\sqrt{2}$ or $x=3 \sqrt{2}$.