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) 

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.
