Find the greatest integer $N$ such that... 
Find the greatest integer $N$ such that $N<\dfrac{1}{\sqrt{33+\sqrt{128}}+\sqrt{2}-8}$.

The way I did it is this: first, I rewrote the biggest square root as $\sqrt{1+2*16+8\sqrt{2}}$. Then I made the substitution $x=\sqrt{2}$, so this became $\sqrt{16x^2+8x+1}$, which factors to $\sqrt{(4x+1)^2}$ which is equal to $4x+1$. Then I substituted back: $\dfrac{1}{4\sqrt{2}+1+\sqrt {2}-8}$, or $\dfrac 1{5 \sqrt{2}-7}$. This is where I have a problem. I am able to find the answer because I know that $1.4<\sqrt 2 <1.5$. However, is there any way to find this answer without knowing this? Thanks!

Picking up from where you left off, we have:
$$ \begin{align*}
N &= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \right\rfloor\\
&= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \cdot \dfrac{5\sqrt{2}+7}{5\sqrt{2}+7} \right\rfloor\\
&= \left\lfloor \dfrac{5\sqrt{2}+7}{50-49} \right\rfloor\\
&= \lfloor 5\sqrt{2}+7 \rfloor\\
&= \lfloor \sqrt{50}+7 \rfloor\\
&= \lfloor \sqrt{50} \rfloor+7  \\
&= \sqrt{49}+7 \\
&= 14 \\
\end{align*}$$
