Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.

Solution:

 Given equation,

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

${{x}^{2}}-2\times x\times 2\sqrt{2}+{{\left( 2\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{2} \right)}^{2}}+6=0$

${{\left( x-2\sqrt{2} \right)}^{2}}={{\left( 2\sqrt{2} \right)}^{2}}-6$

${{\left( x-2\sqrt{2} \right)}^{2}}=\left( 4\times 2 \right)-6=8-6$

${{\left( x-2\sqrt{2} \right)}^{2}}=2$

$\left( x-2\sqrt{2} \right)=\pm \sqrt{2}$

$\left( x-2\sqrt{2} \right)=\sqrt{2}orx=-\sqrt{2}+2\sqrt{2}$

$x=\sqrt{2}+2\sqrt{2}orx=-\sqrt{2}+2\sqrt{2}$

$\Rightarrow x=3\sqrt{2}oex=\sqrt{2}$

Thus, the roots of the given quadratic equation are $x=3\sqrt{2}$ and $x=\sqrt{2}$