<s>[INST] Finding factors of second order complex polynomial. It concerns with finding roots of complex polynomial:
$x^2+(i-1)x+(2+i)$. One way is to find roots by factoring, as:$(x-i)(x -(1-2i))=(x-i)(x-1+2i)=(x^2 -x +ix +i +2)=x^2 +x(i-1)+(i+2)$ 
But, this is a guess game for me, with no formal process to get in this case.
The logical process is to take roots of the quadratic equation, but I hope the quadratic formula does not work for complex coefficients. 

Update -- Based on comments by @WillJagy, the square root is $\pm1\pm3i$, hence the equation is having roots as :$\frac{-(i-1)\pm\sqrt{-6i-8}}2\implies \frac{(1-i)\pm(\pm 1\pm 3i)}2$ can have possible values:  (i) for $1+3i$, get: $x_{11} = 1+i, x_{12} = -2i$
(ii) for $1-3i$, get: $x_{21} = 1-2i, x_{22} = i$
(iii) for $-1+3i$, get: $x_{11} = i, x_{21} = 1-2i$
(iv) for $-1-3i$, get: $x_{11} = -2i, x_{21} = 1+i$
So, the $4$ root pairs should be tried one by one:
(i) $1+3i =>(x-1-i)(x+2i) = x^2 +ix -x -2i +2$, mismatch 
(ii) $1-3i => (x-i)(x-1+2i) = x^2 -x +ix +i +2$, matches original equation 
(iii) same as for (ii),  
(iv) same as for (i)
Why only two root pairs match, and the other don't?

Some answers on mse : 1

Update 2 -- To add to the answer by @Skip, where the fourth quadrant angle was transformed to the first one by taking out $\sqrt{-1} = i$ as common factor. For mixed sign discriminant, let $D = -6 +8i$, then angle cannot be taken\manipulated to be in the first quadrant; & can lie in the 2nd or 3rd quadrant.
In 2nd quad., $\sin$ of negative angle is positive, while in 3rd quad. both $\sin, \cos$ are negative.
 [/INST] Let $z=a+bi, a,b, \in \mathbb{R}$, where $$z^2 = -8-6i$$
$$(a+bi)^2=(-8-6i)$$
$$a^2-b^2+2abi=-8-6i$$
that is we have 
$$a^2-b^2=-8\tag{1}$$
$$ab=-3\tag{2}$$
Substituting $(2)$ into $(1)$,
$$a^2-\frac{9}{a^2}=-8$$
$$a^4+8a^2-9=0$$
$$(a^2-9)(a^2+1)=0$$
$$a=\pm3$$
when $a=3$, from $ab=-3$,$b=-1$.
when $a=-3$, from $ab=-3$, $b=1$ 
There are only two solutions for $z^2=-8-6i$, $z_1 = 3-i$ and $z_2 = -3+i$ and we have $z_2 = -z_1$. That is the solutions to $z^2 = -8-6i$ are $\pm(3-i)$
The solution to the original problem is 
$$x=\frac{(1-i) \pm z_1}{2}$$
You might have solved for $a$ and substitute it into $(1)$ and thought that for each $a$, there are $2$ possible values of $b$'s. However, remember that you have to check all the constraitns, constraint $(2)$ tells us that for each non-zero $a$, there is exactly one $b$.
You shold be expecting $2$ roots counting multiplicity for a quadratic equation from fundamental theorem of algebra.
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