Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$ What I've done so far is:
We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdot i.$
We now need to find the sum $S=x_1+x_2$ and the product $P=x_1\cdot x_2$ to form an equation $x^2-Sx+P=0$ which has the solutions $x_1$ and $x_2$.
\begin{align*}
sum = x_1+x_2&=6m                    \tag{1}\\ 
\iff(r+n\cdot i)+(r-n\cdot i)=2r&=6m \tag{2}\\ 
 \implies r&=3m                      \tag{3}
\end{align*}
The $product = x_1\cdot x_2=(r+n\cdot i)(r-n\cdot i)=r^2+n^2 = (\cdots).\quad $
From here I don't know what relation to find between $\Delta$ and $x_1\cdot x_2.$
I tried: $x_2=\dfrac{-b\pm\sqrt{-36}}{2a}
=\dfrac{-b}{2a}\pm\dfrac{6i}{2a}
=3m\pm\dfrac{3i}{a}\space
\text{ so }\space x_1\cdot x_2
=\dfrac{9m^2+9}{a^2} (\cdots)$

Alternative approach:
Without loss of generality, the equation is 
$x^2 + Bx + C = 0.$
Since $x_1 + x_2 = 6m$, 
and since you must have that 
$(x - x_1) \times (x - x_2) = x^2 + Bx + C$, 
you must have that 
$B = -6m.$

Here, there is some ambiguity involved.  Taking the constraint of (in effect) 
$B^2 - 4C = -36$ 
at face value, you have that
$\displaystyle B^2 + 36 = 4C \implies C = \frac{36m^2  + 36}{4} = 9m^2 + 9.$
Therefore, the quadratic equation is
$\displaystyle x^2 + [-6m]x + \left[9m^2 + 9\right] = 0.$

I mentioned a possible ambiguity.
Generally, for an equation of the form $Ax^2 + Bx + C = 0$, whose solutions are given by 
$\displaystyle \frac{1}{2A} \left[-B \pm \sqrt{B^2 - 4AC}\right]$
it is unclear whether the discriminant should be considered to be
$$ B^2 - 4AC ~~~\text{or}~~~ \frac{B^2 - 4AC}{4A^2}. \tag1 $$
In (1) above, I went with the LHS, on a guess.
