Find the solutions of $3x^2-2\sqrt{6}x+2=0$ by the method of completing the squares when $x$ is an irrational number.

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

Dividing the entire equation, by $3$, we get 

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

First we move the fraction$\frac{2}{3}$ to the right side so that we can make the left side of the equation a perfect square. 

Then we will try to balance the left side by adding the number that will make it a perfect square such that the number will be added on both sides of equation. 

So, $x^2-\frac{2\sqrt{6}}{3}x=-\frac{2}{3}$

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

$\Rightarrow {\left(x-\frac{\sqrt{6}}{3}\right)}^2=-\frac{2}{3}+\frac{6}{9}$  

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

$\Rightarrow x=\frac{\sqrt{6}}{3}$ 

$\Rightarrow x=\frac{\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

$\Rightarrow x=\sqrt{\frac{2}{3}}$ 

Hence, the solution of the equation $3x^2-2\sqrt{6}x+2=0$ is $\sqrt{\frac{\mathbf{2}}{\mathbf{3}}}$.