Terrible equations!

Firstly consider all the values as variables which will make the problem easy to solve.

Let:

$l=\sqrt{\sqrt{5}+2}$

$m=\sqrt{\sqrt{5}-2}$

$n=\sqrt{\sqrt{5}+1}$

So, $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}+\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{2\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}=\frac{l+m}{n}+\frac{l+m}{2n}-\sqrt{3-2\sqrt{2}}$

Taking LCM of $n$ and $2n$ = $2n$ , we get:

$\frac{l+m}{n}+\frac{l+m}{2n}-\sqrt{3-2\sqrt{2}}=\frac{2l+2m+l+m}{2n}-\sqrt{3-2\sqrt{2}}$

$\frac{3l+3m}{2n}-\sqrt{3-2\sqrt{2}}=\frac{3(l+m)}{2n}-\sqrt{3-2\sqrt{2}}$

Putting back the values, we get:

$x=\frac{3(l+m)}{2n}-\sqrt{3-2\sqrt{2}}=\frac{3(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})}{2\sqrt{\sqrt{5}+1}}-\sqrt{2+1-2\sqrt{2}}$

$x=\frac{3(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})}{2\sqrt{\sqrt{5}+1}}-\sqrt{(\sqrt{2}-1)^{2}}$

$x=\frac{3(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})}{2\sqrt{\sqrt{5}+1}}-(\sqrt{2}-1)$

$x=\frac{3(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})}{2\sqrt{\sqrt{5}+1}}-\sqrt{2}+1$

$x+\sqrt{2}-1=\frac{3(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})}{2\sqrt{\sqrt{5}+1}}$

Squaring both sides, we get:

$(x+\sqrt{2}-1)^{2}=(\frac{3(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})}{2\sqrt{\sqrt{5}+1}})^{2}$

$(x+\sqrt{2}-1)^{2}=\frac{(3(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}))^{2}}{(2\sqrt{\sqrt{5}+1})^{2}}$

$(x+\sqrt{2}-1)^{2}=\frac{9(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^{2}}{4(\sqrt{5}+1)}$

$(x+\sqrt{2}-1)^{2}=\frac{9(\sqrt{5}+2+\sqrt{5}-2+2\sqrt{(\sqrt{5}+2)(\sqrt{5}-2)})}{4(\sqrt{5}+1)}$

$(x+\sqrt{2}-1)^{2}=\frac{9(2\sqrt{5}+2\sqrt{5-4})}{4(\sqrt{5}+1)}$

$(x+\sqrt{2}-1)^{2}=\frac{9(2\sqrt{5}+2)}{4(\sqrt{5}+1)}$

$(x+\sqrt{2}-1)^{2}=\frac{18(\sqrt{5}+1)}{4(\sqrt{5}+1)}$

$(x+\sqrt{2}-1)^{2}=\frac{18}{4}$

$(x+\sqrt{2}-1)^{2}=\frac{9}{2}$

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

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

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

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

$x=\frac{\sqrt{2}+2}{2}$

Again, squaring both sides, we get:

$x^{2}=(\frac{\sqrt{2}+2}{2})^{2}$

$x^{2}=\frac{(\sqrt{2}+2)^{2}}{4}$

$x^{2}=\frac{(2+4+4\sqrt{2})}{4}$

$x^{2}=\frac{(6+4\sqrt{2})}{4}$

$x^{2}=\frac{2(3+2\sqrt{2})}{4}$

$x^{2}=\frac{3+2\sqrt{2}}{2}$

Thus, the answer is: $x^{2}=\boxed{\frac{3+2\sqrt{2}}{2}}$