Denesting a square root: $\sqrt{7 + \sqrt{14}}$ Write:
$$\sqrt{7 + \sqrt{14}} = a + b\sqrt{c}$$
Form. $$7 + \sqrt{14} = a^2 + 2ab\sqrt{c} + b^2c$$
$a^2 + b^2c = 7$ and $2ab = 1$, and $c = 14$
But that doesn't seem right as $a, b,$ wont be integers?

Set $r=\sqrt{7+\sqrt{14}}$; then $r^2=7+\sqrt{14}$ and so
$$
14=r^4-14r^2+49
$$
or
$$
r^4-14r^2+35=0
$$
The polynomial $X^4-14X^2+35=0$ is irreducible over the rational numbers by Eisenstein's criterion (with $7$), so the degree of $r$ over the rationals is $4$. A number of the form $a+b\sqrt{c}$ with rational $a,b,c$ has degree $2$ over the rationals.
Therefore you can't find rational $a,b,c$ that satisfy your request.
