Solving a quadratic equation with complex coefficient and one unknown coefficient Consider the equation with $z ∈ ℂ$
$$2z^2 − (3 + 8i)z − (m + 4i) = 0$$
where $m$ is a real constant and such that one of the two solutions is real.
I want to know how to calculate the solutions. I try to use $b^2-4ac$ to solve it, but it is hard to solve it. The unknown coefficient is so tricky
 A: Hint. Let $x$ be the real solution and $z=a+ib$ be the other solution. Then the sum and the product of these solutions can be obtained from the coefficients of the quadratic:
$$x+(a+ib)=\frac{3+8i}{2}\quad \mbox{and}\quad x(a+ib)=−\frac{m + 4}{2}.$$
Hence, since $m,x,a,b\in\mathbb{R}$, after separating real and imaginary parts, we get
$$\begin{cases}
x+a=\frac{3}{2}\\
b=4\\
xa=-\frac{m}{2}\\
xb=-2
\end{cases}$$
Can you take it from here?
A: If $r$ is a real solution, then $2r^2 − (3 + 8)r − (m + 4) = 0$. Hence, $2r^2-3r-m=0$ and $-8r-4=0$.
So, $\displaystyle r=-\frac{1}{2}$ and $m=2$.
The other root is $\displaystyle \frac{3+8i}{2}-\frac{-1}{2}=2+4i$.
A: Hint: by the quadratic formula we get
$$z_{1,2}=\frac{1}{4}(3+8i)\pm\sqrt{\frac{1}{16}(3+8i)^2-\frac{1}{2}(m+4i)}$$
Can you finish?
Second hint: we get the first solution as $$z_1=3/4+1/8\,\sqrt {2\,\sqrt { \left( -55+8\,m \right) ^{2}+6400}-110+16\,
m}+i \left( 2+1/8\,\sqrt {2\,\sqrt { \left( -55+8\,m \right) ^{2}+6400
}+110-16\,m} \right) 
$$
A: HINT
We have
$$2z^2 − (3 + 8)z − (m + 4) = 0\iff z^2 − \frac{3 + 8}2z − \frac{m + 4}2 = 0$$


*

*$z_1+z_2=\frac{3 + 8}2$

*$z_1z_2=− \frac{m + 4}2$
let wlog $z_2=a\in \mathbb{R}$.
