Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential. 
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards

Alternatively, you can use Michael Rozenberg's identities, where this identity applies to the question:
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
We have $a = 28$, $b = 16^2 \cdot 3 = 768$. Applying this identity gives:
$$\sqrt{28+\sqrt{768}}=\sqrt{\frac{28+\sqrt{16}}{2}}+\sqrt{\frac{28-\sqrt{16}}{2}}=\sqrt{16}+\sqrt{12}=4+2\sqrt{3}.$$
Applying the identity once more gives, where $a=4, b=2^2 \cdot 3 = 12$ gives:
$$\sqrt{4+\sqrt{12}}=\sqrt{\frac{4+\sqrt{4^2-12}}{2}}+\sqrt{\frac{4-\sqrt{4^2-12}}{2}}=\sqrt{\frac{4+2}{2}}+\sqrt{\frac{4-2}{2}}=\sqrt{3}+1.$$
