Complex Coefficients and Real roots

Question

Find $m$ which is a real number so that this equation has a real root.

$2z^2-(3+8i)z-(m+4i)=0$

I've tried $b^2-4ac=0 $ but I can only seem to get complex $m$ values, so either I'm missing a key point or just failing with my calculations.

Any help/ideas would really be appreciated.

Thanks :)

Answer 1

HINT:

If $z$ has to be real and as $m$ is also real,

equating the real & the imaginary parts

$$2z^2-3z-m=0$$

and $$8z+4=0$$