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) 

We have that 
$$a^2-b^2=10 \quad \textrm{and} \quad 2ab=-4\sqrt 6$$
Now, squaring both equalities and addem up we get
$$(a^2+b^2)^2=(a^2-b^2)^2 +(2ab)^2=10^2+(-4\sqrt 6)^2 =196$$
$$\Rightarrow \quad a^2+b^2=14$$
and using again the first equality we obtain
$$a^2=12 \quad \textrm{and} \quad b^2=2$$
or
$$a=\pm 2\sqrt 3 \quad \textrm{and} \quad b=\pm \sqrt 2$$
but $ab<0$ (by the second equality), that is, $a$ and $b$ are the opposite sign. Thus, the solutions are given by
$$a= 2\sqrt 3 \quad \textrm{and} \quad b=- \sqrt 2$$
or well
$$a=-2\sqrt 3 \quad \textrm{and} \quad b=\sqrt 2$$
