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

To find the fourth root, we need to use Bill Dubuque's denesting algorithm for square roots twice.
First time round, we can find that $\sqrt{\text{norm}} = \sqrt{28^2 - 3 \cdot 16^2} = \sqrt{16} = 4$. Subtracting out the norm gives $24 + 16 \sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{2 \cdot 24} = 4 \sqrt{3}$, we get $\frac{8 \sqrt{3}\sqrt 3}{4 \sqrt 3} + \frac{16 \sqrt3}{4 \sqrt3} = 2\sqrt{3}+4 = 4 + 2\sqrt{3}$.
Second time round, we can find that $\sqrt{\text{norm}} = \sqrt{4^2 - 3 \cdot 2^2  -8} = \sqrt{4} = 2$. Subtracting out the norm gives $2 + 2\sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{4} = 2$, you get $1 + \sqrt{3}$.
