<s>[INST] Evaluating $\sqrt{a\pm bi\sqrt c}$ I recently encountered this problem
$$\sqrt{10-4i\sqrt{6}}$$
To witch I set the solution equal to $a+bi$ squaring both sides leaves
$${10-4i\sqrt{6}}=a^2-b^2+2abi$$
Obviously $a^2-b^2=10$ and $2abi=-4i\sqrt{6}$, using geuss and check, the solution is $a=\sqrt12, b=\sqrt2$
But I was wondering if there is a faster way to solve these types of problems or a method that doesn't involve guess and check (since it can get tedious at times) for the basic form 
$$\sqrt{a\pm bi\sqrt c}=x+yi$$ Where you're solving for $x$ and $y$.
I've attempted but failed since it gets pretty messy. All formula will be very much appreciated. (not all equations of that form can reduce) 
 [/INST] One thing right off the bat.  Every non-zero complex number will have two square roots so there wont be just $x + yi$ there will also be $-x -yi$
" using geuss and check"  Why use guess and check when you can use the quadratic formula?
$a^2 - b^2 = 10$ and $2abi = -4\sqrt 6i$
$a=\frac{-2\sqrt 6}b$
$\frac {4*6}{b^2} - b^2 = 10$
$24 - b^4 = 10b^2$
$b^4 + 10b^2 - 24 = 0$
$b^2 = \frac {-10 \pm \sqrt{100+96}}2= -5 \pm \sqrt{25+ 24}=-5\pm 7$.
As $b$ is presumed to be real $b^2 = 2$ and $b=\pm \sqrt 2$
And $a=\frac{-2\sqrt 6}b= \mp \frac {2\sqrt 6}{\sqrt{2}}=\pm 2\sqrt 3(= \pm \sqrt {12})$
So $\sqrt{10-4\sqrt 6i} = \pm(-2\sqrt 3 + \sqrt 2i)$
No guessing.  No checking.
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