2z²-(3+8i)z-(m+4i)=0, and if one of the roots z₁=x where x is real, then
2x²-3x-8ix-m-4i=0. Therefore -8x-4=0 because there is no imaginary component remaining, making x=-½ and z₁=-½. The quadratic in x becomes 2/4+3/2-m=0, making m=2
Rewrite the equation: z²-½(3+8i)-(1+2i)=0.
The sum of the roots z₁+z₂=½(3+8i) so z₂=½(3+8i)+½=2+4i.
The product of the roots is -1-2i=-½(2+4i)=z₁z₂, confirming the solution.
The roots are -½ and 2+4i.