Since the quadratic equation has complex coefficients, it is not the case that the two solutions are complex conjugates of one another. However, the two solutions do lie diametrically opposite one another on a circle of radius $ \ \rho \ $ centered at a complex number $ \ z_0 \ $ . The diameter makes an angle $ \ \theta \ $ to the "real" axis, so the two solutions can be expressed as $ \ z \ = \ z_0 \ \pm \ \rho \ cis \ \theta \ $ or $ \ z \ = \ z_0 \ \pm \ \rho \ e^{ \ i \theta} \ $ . (I am expressing it this way to demonstrate how this might be done if you aren't used to taking square-roots of complex numbers.)
If we work with a quadratic polynomial with a leading coefficient of 1 , it may be factored as $ \ ( z \ - \ z_0 \ - \ \rho \ e^{ \ i \theta} ) \ ( z \ - \ z_0 \ + \ \rho \ e^{ \ i \theta} ) \ $ . Multiplying this out for the quadratic equation produces
$$ z^2 \ - \ 2 \ z_0 \ z \ + \ ( \ z_0^2 \ - \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ ) \ = \ 0 \ \ . $$
Martin Sleziak shows the monic polynomial equation we wish to solve, so we can compare coefficients with $ \ z^2 \ -(1+3i) \ z \ + \ (-2+i) \ = \ 0 \ $ . We have immediately that $ \ z_0 \ = \ \frac{1 \ + \ 3i}{2} \ $ . [In fact, what we are doing is essentially what "completing the square" provides us.] The square of this number is $ \ z_0^2 \ = \ -2 \ + \ \frac{3}{2}i \ $ , allowing us to establish the equation
$$ z_0^2 \ - \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ = \ -2 \ + \ i $$
$$ \Rightarrow \ \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ = \ ( -2 \ + \ \frac{3}{2}i ) \ - \ ( -2 \ + \ i ) \ = \ \frac{1}{2}i \ \ . $$
The modulus of $ \ e^{ \ i \cdot 2 \theta} \ $ is 1 , so we can write $ \ | \ \rho^2 \ e^{ \ i \cdot 2 \theta} \ | \ = \ | \ \rho^2 \ | \ = \ | \ \frac{1}{2}i \ | \ = \ \frac{1}{2} \ $ . Inserting this in the earlier equation leaves us with $ \ \frac{1}{2} \ \cdot \ \ e^{ \ i \cdot 2 \theta} \ = \ \frac{1}{2}i \ \ \Rightarrow \ \ e^{ \ i \cdot 2 \theta} \ = \ i \ $ .
Now we have reached the part where we are extracting the square-root. "Euler's Identity" gives us $ \ e^{ \ i \cdot 2 \theta} \ = \ \cos \ 2 \theta \ + \ i \ \sin \ 2\theta \ = \ 0 \ + \ 1 \cdot i \ $ . It remains to solve the simultaneous trigonometric equations
$$ \cos \ 2 \theta \ = \ 0 \ \ \text{and} \ \ \sin \ 2 \theta \ = \ 1 \ \ \Rightarrow \ \ 2 \theta \ = \ \frac{\pi}{2} \ + \ 2k \cdot \pi \ \ \Rightarrow \ \ \theta \ = \ \frac{\pi}{4} \ + \ k \cdot \pi \ \ . $$
So the two angles involved are $ \ \frac{\pi}{4} \ \ \text{and} \ \ \frac{5 \pi}{4} \ $ (and, redundantly for our purpose, all the other "angle-names" in these directions). Bringing together our results, the two solutions to the complex quadratic equation are
$$ \frac{1 \ + \ 3i}{2} \ + \ \rho \ (\cos \ \frac{\pi}{4} \ + \ i \ \sin \ \frac{\pi}{4} ) \ \ , \ \ \frac{1 \ + \ 3i}{2} \ + \ \rho \ (\cos \ \frac{5 \pi}{4} \ + \ i \ \sin \ \frac{5 \pi}{4} ) $$
$$ = \ \frac{1 \ + \ 3i}{2} \ + \ \frac{1}{\sqrt{2}} \ ( \frac{\sqrt{2}}{2} \ + \ i \ \frac{\sqrt{2}}{2} ) \ \ , \ \ \frac{1 \ + \ 3i}{2} \ + \ \frac{1}{\sqrt{2}} \ ( -\frac{\sqrt{2}}{2} \ - \ i \ \frac{\sqrt{2}}{2} ) $$
$$ = \ \left(\frac{1 }{2} \ + \ \frac{1}{ 2 } \right) \ + \ i \ \left( \frac{3}{2} \ + \ \frac{1}{2} \right) \ \ , \ \ \left(\frac{1 }{2} \ - \ \frac{1}{ 2 } \right) \ + \ i \ \left( \frac{3}{2} \ - \ \frac{1}{2} \right) $$
$$ = \ \ 1 \ + \ 2i \ \ , \ \ 0 \ + \ i \ \ , $$
agreeing with the solutions shown by Vinod Kumar Punia and by WA. (I will say that things work out rather tidily for this equation; we would have somewhat more work to do in general to solve the pair of trigonometric equations, but it can still be managed "by hand" with the real and imaginary parts of the coefficients being integers or even rational numbers.)
