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: $x_2=\dfrac{-b\pm\sqrt{-36}}{2a}
=\dfrac{-b}{2a}\pm\dfrac{6i}{2a}
=3m\pm\dfrac{3i}{a}\space
\text{ so }\space x_1\cdot x_2
=\dfrac{9m^2+9}{a^2} (\cdots)$
 A: 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.
A: 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.
A: \begin{align*}
b^2-4ac=-36\implies 4ac\ge36\implies ac&\ge9\tag{1}\\
 \\
x_1+x_2=
\bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg)+
\bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg)=
\dfrac{-b}{a}&=6m\tag{2}\\
\\
x_1\cdot x_2=
\bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg) 
\bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg)
= \dfrac{b^2+36}{4 a^2}&  \tag{3}
\end{align*}
From $(1)$ we could say $\space x^2+36=0\space$ but
$(2)\longrightarrow\dfrac{-0}{a}=6m\implies m=0.$
The first "candidates" for $\space (a, b)\space$ that will yield $\space -36\space$ are $a=1, b=2$
\begin{align*}
\dfrac{-b}{a}=\dfrac{-2}{1}
\implies a=1, b=2, 4ac=40\\
\implies 2^2-4(1)(10)=-36\\
\implies c=\dfrac{2^2+36}{4 (1^2)}=10\\
\\
\implies 
x^2+2x+10=0
\\ \implies
 x_1= -1+3i\quad  x_2=-2-3i\\ \\
x_1\cdot x_2=( -1+3i)(-1-3i)=10 \\
\dfrac{-2}{1}=6m\implies m=\dfrac{-1}{3}
\end{align*}
These last equations are in no particular order but I think we have the quadratic equation we seek, i.e.
$\quad x^2+2x+10=0$
