What is the square root of $3 + 2\sqrt{10}i$? I need to compute the square root of $3 + 2\sqrt{10}i$.

I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this: 

$$
  \sqrt{3 + 2\sqrt{10}i} = x + iy
    \quad \longrightarrow \quad
  3 + 2\sqrt{10}i = x^2 - y^2 +2xyi
$$
and so forth, but my answer isn't correct.
As I explained here. there is  a very simple formula for denesting such radicals, namely

Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $

$\ 3+2\sqrt{-10}\ $ has norm $= 49.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = 7\,\ $ yields $\ {-}4+2\sqrt{-10}\:$

with $\, {\rm\ \sqrt{trace}}\, =\, 2\sqrt{-2}.\ \ \ \rm \color{brown}{Dividing\ this\ out}\ $ of the above we obtain $\,\ \sqrt{-2} + \sqrt 5$ 

Checking it we find  $\,\ (\sqrt{-2} + \sqrt 5)^2 =\, -2+5 + 2\sqrt{-2}\sqrt 5\, =\, 3+ 2\sqrt{-10}$

Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
You want $z=a+bi$ where $a,b\in\mathbb{R}$ such that

$$z^2=a^2-b^2+2abi=3+2\sqrt{10}i.$$

Comparing coefficients, you need $a^2-b^2=3$ and $ab=\sqrt{10}$. So $a=\sqrt{5}$ and $b=\sqrt{2}$, or $a=-\sqrt{5}$ and $b=-\sqrt{2}$. Thus the square root of $3+2\sqrt{10}$:

$$\pm(\sqrt{5}+\sqrt{2}i).$$