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: $x_2=\dfrac{-b\pm\sqrt{-36}}{2a}
=\dfrac{-b}{2a}\pm\dfrac{6i}{2a}
=3m\pm\dfrac{3i}{a}\space
\text{ so }\space x_1\cdot x_2
=\dfrac{9m^2+9}{a^2} (\cdots)$

You already concluded from the negative discriminant that the zeroes of the quadratic polynomial must be a "conjugate pair" $ \ \alpha \ + \ \beta·i \ \ . $  The difference of the zeroes is
$$ x_1 \ - \ x_2 \  \ = \ \ \frac{\sqrt{\Delta}}{a} \ \ \Rightarrow \ \ 2·\beta·i \ \ = \ \ \frac{\sqrt{-36}}{a} \ \ = \ \ \frac{6i}{a} \ \ \Rightarrow \ \ \beta \ = \ \frac{3}{a} \ \ . $$
The sum of the zeroes is given as
$$ x_1 \ + \ x_2 \ \ = \ \ 2·\alpha \ \ = \ \ 6m \ \ \Rightarrow \ \ \alpha \ = \ 3m \ \  $$
(as you observed).
If we choose to construct the corresponding monic polynomial, we would set $ \ a = 1 \ \ , $ producing the conjugate pair of zeroes $ \ x_{1,2} \ = \ 3m \ \pm \ 3i \ \ . $  Multiplying out the factors $ \ (x \ - \ 3m \ - \ 3i) \ · \ (x \ - \ 3m \ + \ 3i) \ \ $ gives us the polynomial shown in  user2661923's answer.
