METHOD 1
Multiply through by 1-i: (1+1)x^2-ix(1-i)=6(1-i)
2x^2-ix+i^2x=6(1-i); 2x^2-ix-x=6(1-i)
x^2-x(i+1)/2=3(1-i) (dividing through by 2)
x^2-x(i+1)/2+(i+1)^2/16=3-3i+(i+1)^2/16 (completing the square and balancing the equation)
(x-(i+1)/4)^2=3-3i+(-1+2i+1)/16=3-3i+i/8=3-23i/8
x-(i+1)/4=+sqrt(3-23i/8)
Consider, where A and B are real: 3-23i/8=(A+iB)^2=A^2+2ABi-B^2; A^2-B^2=3 2AB=-23/8, by equating real and imaginary components. We can solve for real A and B:
B=-23/16A and A^2-529/256A^2=3; 256A^4-529=768A^2
256A^4-768A^2=529; 256(A^4-3A+9/4)=529+576; 256(A^2-3/2)^2=1105
16(A^2-3/2)=+sqrt(1105); A^2-3/2=+sqrt(1105)/16; A^2=3/2+sqrt(1105)/16: A^2=3.5776 approx., real A=1.8915.
B=-23/16A=-0.7600 approx. So sqrt(3-5i/2)=1.8915-0.7600i.
and x=(i+1)/4+(1.8915-0.7600i)=2.1415-0.5100i or -1.6415+1.0100i.
METHOD 2
(1+i)x^2-ix-6=0
Use the quadratic formula to solve:
x=(i+sqrt(-1+24(1+i))/2(1+i)=(i+sqrt(23+24i))/2(1+i)
Let (A+iB)^2=23+24i; A^2+2ABi-B^2=23+24i; A^2-B^2=23; 2AB=24, AB=12, B=12/A.
A^2-144/A^2=23; A^4-23A^2-144=0, A^2=(23+sqrt(529+576))/2=(23+sqrt(1105))/2=28.12077 and A=5.3029 and B=2.2629.
x=(i+(A+iB))/2(1+i))=(+A+i(1+B))/2(1+i)
Multiply top and bottom by 1-i:
x=(+A+i(1+B)-i(+A+i(1+B))/4=(+A+i+iB-i(+A)+1+B)/4=(+A+1+B)+i(1+B-(+A)))/4 (real and imaginary split).
This gives us two values of x: 2.1415-0.5100i or -1.6415+1.0100i.
The two methods produce the same solution.