Squaring the fraction gives
$$\frac{7+\sqrt5}{7-\sqrt5}=\frac{1}{44}(7+\sqrt 5)^2$$
so by taking the square root we find
$$\frac{7+\sqrt 5}{2\sqrt{11}}$$
Your work is just fine: you've shown you know that $\dfrac 12 = \sqrt{\dfrac{1}{4}},\;\;$ and that for any $x,y\in \mathbb{R^+\cup \{0\}},\;\;\sqrt x \cdot \sqrt y=\sqrt{xy}$.
The negative sign outside of the radicand has no impact on your operations: since the operations between terms is strictly multiplication, we can operate (multiply) as if the positive terms are entirely contained within parentheses, all of which is then multiplied by $-1$:
Note: the parentheses are used for illustration only: to make explicit that your computation is indeed correct. But, in fact, parentheses are not necessary.
Best Answer
Squaring the fraction gives $$\frac{7+\sqrt5}{7-\sqrt5}=\frac{1}{44}(7+\sqrt 5)^2$$ so by taking the square root we find $$\frac{7+\sqrt 5}{2\sqrt{11}}$$