Question: > 
> Show that there exists $m \in \mathbb{R}$
>  such that $2z^2 - (3 - 3i)z - (m-9i) = 0$ has a real solution.
> 
> 
> 

Using $b^2 - 4ac \geq 0$ I was able to arrive at $8m - 90i \geq 0$ but am not sure how to proceed.

Answer: 
Let's suppose there's a real solution $z$ to the equation,
$$2z^2 - (3-3i)z - (m-9i) = 0.$$
Equating real and imaginary parts, we must have
$$2z^2 - 3z - m = 0,$$
as well as
$$3z + 9 = 0.$$
The second equation tells us that *if* such a solution $z$ were to exist, then it must be $z=-3$. Substituting $z=-3$ into the first equation tells us that the corresponding $m$ must be $m=27$.

We can now directly verify that $(z,m)=(-3,27)$ is indeed the unique real solution:
$$2(-3)^2 - (3-3i)(-3) - (27-9i) = 0.$$
Using $b^2-4ac \geq 0$ to show there is real solution works in the Real number domain and not in the Complex number domain because for instance you cant say that $1+i > 1-i$

From Wikipedia : it is impossible to define any relation $\leq$ so that ${\displaystyle (\mathbb {C} ,+,\times ,\leq )} $ becomes an ordered field.

Now back to the question, the answers to this equations are $z = \frac{3-3i \pm \sqrt{(3-3i)^2+4 \*2\* (m-9i)}}{4}$, expanding the expression inside the square root, we get that $z = \frac{3-3i \pm \sqrt{8m -90i}}{4}$, we want that $Im(3-3i \pm \sqrt{8m-90i}) = 0 $ where $Im(z)$ is the imaginary part of $z$ which means that $\sqrt{8m-90i} = x \pm 3i$ where $x$ is Real number squaring both sides, we get that $8m-90i = x^2 \pm 6x i -9$, Now if they are equal then sure that the imaginary part is also equal and the imaginary part is $-90 i = \pm 6 x i$ because $8m$ is Real since $m$ is real so $Im(8m)=0$ and $x^2-9$ is Real because $x$ is real so $Im(x^2-9)=0$ leaving $-90i = \pm 6x i$ dividing by $i$ and solving for $x$ we get that $x= \pm 15$ , trying to find $m$ we get that $8m = x^2-9 = (\pm 15)^2-9 = 216$ so $m=27$. 

So the solutions when $m=27$ is $z=-3$ or $ z=\frac{9}{2}-\frac{3 i}{2}$

which concludes the proof.