square root of $\frac{2+\sqrt{3}}{4}$ 
Find the square root of $\frac{2+\sqrt{3}}{4}$.

Attempt:
I'm thinking of equating it to $a + bi$. And then finding the root. Is that right?

For convenience we will work with $4(2+\sqrt3)$ to avoid fractions.
Let us search a solution of the form $a+b\sqrt3$, giving
$$(a+b\sqrt3)^2=a^2+3b^2+2ab\sqrt3=8+4\sqrt3.$$
By identifying $2ab$ and $4$, we have $b=\dfrac2a$ and
$$a^2+\frac{12}{a^2}+4\sqrt3=8+4\sqrt3.$$
By inspection (or resolution of a quadratic equation), $a^2=2$, so that
$$\left(\sqrt2+\frac2{\sqrt2}\sqrt3\right)^2=8+4\sqrt3,$$
$$\left(\frac{\sqrt2+\sqrt6}4\right)^2=\frac{2+\sqrt3}4.$$
