How do I go about simplifying this complex radical? I'm having a difficult time trying to simplify the radical below. When I type the radical to the left into my calculator, I arrive at the correct answer to the right. But I cannot seem to figure out how to simplify the radical on the left by hand. I've googled the heck out of it, but perhaps I'm using an incorrect search phrase, because I cannot find a similar example. Can someone please either point in the right direction or give me a step by step system for simplifying the radical on the left to arrive to the solution on the right?
radical image:
$$\sqrt{\frac{1+\frac{\sqrt{39}}{8}}{2}}=\frac{\sqrt{26}+\sqrt 6}{8}$$

Method One:
This is another technique for simplifying nested radicals of this form if they can be simplified. It is based on the following binomial formula $a^2+2ab+b^2$. First let's put the nested radical in this form $\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}$. Set $$a^2+b^2=\frac{1}{2}$$ and $$2ab=\frac{\sqrt{39}}{16}$$ corresponding to the each real number underneath the radical sign. The second equation results in $b=\frac{\sqrt{39}}{32a}$.   Substitute b into the first equation to arrive at the following equation $a^2+(\frac{\sqrt{39}}{32a})^2=\frac{1}{2}$; $a^2+\frac{39}{1,024a^2}=\frac{1}{2}$. Multiply all terms by $a^2$ and set the equation equal to zero. So, $$a^4-\frac{1}{2}a^2+\frac{39}{1,024}=0$$ Use the quadratic formula to arrive at $a_1^2=\frac{26}{64}$ and $a_2^2=\frac{6}{64}$ or $a_1=\pm\frac{\sqrt26}{8}$ and $a_2=\pm\frac{\sqrt6}{8}$. If we add the two positive roots $a_1$ and $a_2$ and we can see that $$\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}=\frac{\sqrt{26}+\sqrt 6}{8}$$ The equality $\sqrt{\frac{1}{2}-\frac{\sqrt{39}}{16}}$=$\frac{\sqrt{26}-\sqrt 6}{8}$ is computed the same way. If the result of these computations do not denest the nested radical nicely the result will be even for nested than the original nested radical. Nevertheless, the result will still be equal to the original nested radical.
Method Two:
Another similar method where we use the quadratic equation but only to degree two not four is as follows. Set the nested radical as the sum of two square roots so that $$\sqrt{\frac{1}{2}+\frac{\sqrt{39}}{16}}=(\sqrt{a}+\sqrt{b})$$ Then square both sides so that $$\frac{1}{2}+\frac{\sqrt{39}}{16}=a+2\sqrt{a}\sqrt{b}+b$$ Set (1) $$a+b=\frac{1}{2}$$ and set (2) $$2\sqrt{a}\sqrt{b}=\frac{\sqrt{39}}{16}$$   Square both sides of (2) to get $4ab= \frac{{39}}{256}$ and solve for $b$ to get $b=\frac{39}{1,024a}$. Putting $b$ into (1) gives $a+\frac{39}{1,024a}=\frac{1}{2}$. Multiply all terms by $a$ and convert to the quadratic equation $$a^2-\frac{1}{2}a+\frac{39}{1,024}=0$$ Solving the quadratic gives $a=\frac{26}{64}$or $a=\frac{6}{64}$. As a result at the same time $b=\frac{26}{64}$or $b=\frac{6}{64}$. Replacing $a$ and $b$ in the sum or square roots formula above gives $\frac{\sqrt{26}+\sqrt 6}{8}$.
