How to equalize correctly? If i have this number: $2 \sqrt{2-\sqrt{3}}$ and i want to find some $x,y$ nonzero real numbers such that $2\sqrt{2-\sqrt{3}} = \sqrt{x} + \sqrt{y}$ And for that, i do this: $(2 \sqrt{2-\sqrt{3}})^2 = x + 2\sqrt{xy} + y$ $4(2-\sqrt{3})=(x+y)+2\sqrt{xy}$ $(8)+(-4\sqrt{3})=(x+y)+(2\sqrt{xy})$ Then: $i) 8 = x+y$, $ii)-4\sqrt{3} = 2\sqrt{xy} => -2\sqrt{3}=\sqrt{xy}$ $ii) = (-2\sqrt{3})^2= (\sqrt{xy})^2 => 4\cdot 3=xy , x = 12/y$ And solving the equation $y^2-8y+12=0$ gives $y_{1,2} = \\{6,2\\}$ But $2\sqrt{2-\sqrt{3}} \neq \sqrt{6} + \sqrt{2}$ I know that the correct value must be $\sqrt{6} - \sqrt{2}$ but my result is different. What is wrong with my development?

You should assume a binomial of the form whose root you desire. In this case, you should have supposed the root is $$\sqrt x -\sqrt y$$ instead.

To be specific, the problem in the above calculation is in your step ii, where you set $$-\sqrt{12}=\sqrt{xy}.$$ But this is impossible if you're dealing only with real numbers. It seems you need to note that the symbol $\sqrt{}$ denotes a function which, by definition, assumes nonnegative values. Thus, you can see that your equation is false, for it says a negative number is equal to a nonnegative one. That's a contradiction.