How to find root of the complex number?

When finding nth root of the complex number $z=\cos{x}+i\sin{x}$, we can use known De Moivre's formula:

$z^\frac1n=[r(\cos{x}+i\sin{x})]^\frac1n=r^\frac1n\left[\cos{\left(\frac{x+2k\pi}n\right)}+i\sin{\left(\frac{x+2k\pi}n\right)}\right]$

My task is to find $\sqrt{3+2i}$ without using De Moivre's formula, and my question is how to do that?

Thank you for your answers.

Here's a technique that I can only recommend because you asked this question. Suppose that $$\sqrt{3+2i} = a + b\,i$$ or, equivalently, $$3+2\,i = (a+b\,i)^2 = (a^2-b^2) + (2ab)i.$$ Equating real and imaginary parts, we get \begin{align} a^2-b^2 &= 3 \\ 2ab &= 2. \end{align} Thus, you can find $a$ and $b$ by solving this system. From the first equation, we have $a^2=b^2+3$; you can plug that back into the second equation to get a quartic that can, in principle, be solved using Tartaglia's method. Applying Mathematica to this system, I found that $$a+bi= \pm \left(\sqrt{\frac{1}{2} \left(3+\sqrt{13}\right)}+\frac{1}{2} i \left(\sqrt{13}-3\right) \sqrt{\frac{1}{2} \left(3+\sqrt{13}\right)}\right).$$

Note that if the root is $a+bi$ then $a^2-b^2=3$ and $2ab=2$

Then $a^2b^2=1$ so that $(a^2-b^2)^2+4a^2b^2=(a^2+b^2)^2=13$ and $a^2+b^2=\sqrt {13}$

Then it is easy to solve for $a^2$ and $b^2$, bearing in mind that $a$ and $b$ are real. Then you get $a$ and $b$, and fix the signs of the square roots to make the product come out with the right sign.