Simplify nested radical

$\sqrt{9-2\sqrt{14}}=\sqrt{7-2\sqrt{14}+2}$

$=\sqrt{(\sqrt{7})^{2}-2\sqrt{7}\sqrt{2}+(\sqrt{2})^{2}}=\sqrt{(\sqrt{7}-\sqrt{2})^{2}}=\sqrt{7}-\sqrt{2}$

$a=7$ and $b=2$ , so $a-b=\boxed{5}$

For those who cannot see the perfect square:

$\sqrt{9-2\sqrt{14}} = \sqrt{a} - \sqrt{b}\\ 9-2\sqrt{14} = a+b - 2\sqrt{ab}$

By comparison:

$a+b=9 \quad\quad -2\sqrt{14} =-2\sqrt{ab}$

Use the second equation:

$-2\sqrt{14} = -2\sqrt{ab}\\ ab = 14\\ b=\dfrac{14}{a}$

Substitute into the first equation:

$a+\dfrac{14}{a} = 9\\ a^2 - 9a + 14 = 0\\ (a-7)(a-2) = 0\\ a=7,\;2$

When $a=7,\;b= \dfrac{14}{7} = 2$

When $a=2,\;b=\dfrac{14}{2} = 7$

Now, notice that $\sqrt{9-2\sqrt{14}} > 0$ , therefore $\sqrt{a} > \sqrt{b} \implies a>b$

Therefore, $a=7,\;b=2$ and $a-b = 7-2 = \boxed{5}$

$\Large{\sqrt{9-2\sqrt{14}}\\=\sqrt{(\sqrt 7)^2+(\sqrt{2})^2-2\sqrt 7\cdot\sqrt 2}}$ $\Large =\sqrt 7-\sqrt 2$ Hence, $\Large 7-2=\boxed 5$ .

Otherwise we can square both side and combare to get $\large a+b=9$ and $\large ab=14$ which on solving gives $\large a=7,b=2$ .