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

Question

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 1

As commenters pointed out, $80$ should be $81$. The radicals simplify as follows: $$ 3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{81}{16}} = 3\sqrt{2}- \sqrt{16} \sqrt{2} +\frac94 = (3-4)\sqrt{2}+\frac94 = \frac94-\sqrt{2} $$

Answer 2

Assuming you meant $81$: $$ \begin {align*} 3\sqrt2 - \sqrt{32} + \sqrt{\dfrac{81}{16}} &= 3\sqrt{2}- \sqrt{16} \sqrt{2} +\frac94 \\&= (3-4)\sqrt{2}+\frac94 \\&= \frac94-\sqrt{2}. \end {align*} $$