The quadratic equation is #3x^2+(1+2i)x+1-i=0#
now its discriminant is #(1+2i)^2-4*3*(1-i)#
= #1-4+4i-12+12i=-15+16i#
Hence using quadratic formula its roots are
#x=(-(1+2i)+-sqrt(-15+16i))/6#
Let #sqrt(-15+16i)=a+bi#, then squaring
#-15+16i=a^2-b^2+2abi#
i.e. #a^2-b^2=-15# and #2ab=16#
and #(a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2=225+256=481#
i.e. #a^2+b^2=sqrt481#
and #a^2=(sqrt481-15)/2# and #b^2=(sqrt481+15)/2#
i.e #sqrt(-15+16i)=sqrt((sqrt481+15)/2)+isqrt((sqrt481+15)/2)#
and putting this we get
#x=(-(1+2i)+-(sqrt((sqrt481+15)/2)+isqrt((sqrt481+15)/2)))/6#
= #[-1+-sqrt((sqrt481+15)/2)]/6+(-2+-sqrt((sqrt481+15)/2))/6i#
Note that if we choose #+# in real part, we choose #+# in imaginary part too and if we choose #-# in real part, we choose #-# in imaginary part too.
