A hard square root question This is my first question on StackExchange. So my question is:
If $$x = \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{\sqrt{\sqrt5 + 1}} + \frac{\sqrt{\sqrt5 +2}+\sqrt{\sqrt5-2}}{2\sqrt{\sqrt5 + 1}} - \sqrt{3-2\sqrt2}$$
What is the value of $x^2$? 
If someone can also tell me how to input mathematical functions on the StackExchange it would be appreciated.

Letting
$$\frac{\sqrt{\sqrt 5+2}+\sqrt{\sqrt 5-2}}{\sqrt{\sqrt 5+1}}=\alpha,$$
we have
$$\alpha^2=\frac{(\sqrt 5+2)+(\sqrt 5-2)+2\sqrt{(\sqrt 5+2)(\sqrt 5-2)}}{\sqrt 5+1}=\frac{2(\sqrt 5+1)}{\sqrt 5+1}=2\Rightarrow \alpha=\sqrt 2.$$
Hence, we have
$$x=\alpha+\frac{\alpha}{2}-\sqrt{(\sqrt 2-1)^2}=\frac{3}{2}\alpha-(\sqrt 2-1)=\frac{\sqrt 2}{2}+1\Rightarrow x^2=\frac{3}{2}+\sqrt 2.$$
