algebra precalculus - How to simplify this expression with radicals? $3sqrt2 - sqrt{32} + sqrt{frac{80}{16}}$

I don't understand how I could calculate this:
$3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{80}{16}}$

My answer is $-\sqrt2 + \sqrt5$, but the real answer should be $\dfrac{9-4\sqrt2}{4}$.

Answer

As commenters pointed out, $80$ should be $81$. The radicals simplify as follows:

$$3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{81}{16}} = 3\sqrt{2}- \sqrt{16} \sqrt{2} +\frac94 = (3-4)\sqrt{2}+\frac94 = \frac94-\sqrt{2}$$