Solve the equation $x^2-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$ by the method of completing the square.

Given equation: $x^2-(\sqrt{3}+1)x+\sqrt{3}=0$. 

We will use the completing square method.

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 the equation. 

${\left(x\right)}^2-2\times \left(x\right)\times \left(\frac{\sqrt{3}+1}{2}\right)+\sqrt{3}=0$ 

$\Rightarrow {\left(x\right)}^2-2\times \left(x\right)\times \left(\frac{\sqrt{3}+1}{2}\right)+{\left(\frac{\sqrt{3}+1}{2}\right)}^2-{\left(\frac{\sqrt{3}+1}{2}\right)}^2+\sqrt{3}=0$

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2+\sqrt{3}-{\left(\frac{\sqrt{3}+1}{2}\right)}^2=0$ 

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2+\sqrt{3}-\frac{4+2\sqrt{3}}{4}=0$ 

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2+\sqrt{3}-1-\frac{\sqrt{3}}{2}=0$  

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2=1-\frac{\sqrt{3}}{2}$  

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2=\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{\sqrt{3}}{2}$

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2={\left(\frac{\sqrt{3}}{2}\right)}^2-2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}+{\left(\frac{1}{2}\right)}^2$

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2={\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)}^2$

$\Rightarrow {\left(x-\frac{\sqrt{3}+1}{2}\right)}^2={\left(\frac{\sqrt{3}-1}{2}\right)}^2$

$\Rightarrow x-\frac{\sqrt{3}+1}{2}=\pm \frac{\sqrt{3}-1}{2}$

Thus, $\mathit{x}\mathbf{=}\sqrt{\mathbf{3}}\mathbf{,}\mathbf{ }\mathbf{1}$.