How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$ 
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$

How to change $\sqrt{2+\sqrt{3}}$ into  $\dfrac{\sqrt{6}+\sqrt{2}}{2}$

$ (\sqrt u + \sqrt v)^2 = u+v + 2\sqrt{u\,v }$ 
So at first bring in $2$ in front of the radical sign.
$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}$$
Next find two factors for $3$ whose sum is $4$
They are easy to guess. They are $3$ and $1$; 
(... or else you need to solve another quadratic equation)
$$ \sqrt{\frac{(\sqrt3 +1)^2}{2}}=\sqrt{  \frac{2(\sqrt3 +1)^2}{4}} $$
where the denominator is made to be $4$ so that while square rooting, $2$ would stay there.
Next manipulate numerator to make them  squares of two terms and make denominator also a square
$$\sqrt{  \frac{(\sqrt{2})^2(\sqrt3 +1)^2}{2^2}} $$
that  is now ( cancelling out squares and square roots) and multiplying two terms in the numerator:
$$ \dfrac{\sqrt 6 +\sqrt 2}{2}.$$
