#### Answer

$m=\left\{ -3-i\sqrt{3},-3+i\sqrt{3} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find the non real complex solutions of the given equation, $
-m^2-6m-12=0
,$ use the properties of equality to express the given equation in the form $x^2+bx=c.$ Then complete the square by adding $\left(\dfrac{b}{2} \right)^2$ to both sides. Factor the left side then take the square root (Square Root Property) of both sides. Then use the properties of radicals and use $i=\sqrt{-1}.$ Finally, simplify the radical and isolate the variable.
$\bf{\text{Solution Details:}}$
Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{-m^2-6m-12}{-1}=\dfrac{0}{-1}
\\\\
m^2+6m+12=0
\\\\
m^2+6m=-12
.\end{array}
In the equation above, $b=
6
.$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then
\begin{array}{l}\require{cancel}
\left( \dfrac{6}{2} \right)^2
\\\\=
\left( 3 \right)^2
\\\\=
9
.\end{array}
Adding $\left(\dfrac{b}{2} \right)^2$ to both sides of the equation above to complete the square, the equation becomes
\begin{array}{l}\require{cancel}
m^2+6m+9=-12+9
\\\\
(m+3)^2=-3
.\end{array}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{array}{l}\require{cancel}
m+3=\pm\sqrt{-3}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
m+3=\pm\sqrt{-1}\cdot\sqrt{3}
.\end{array}
Using $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
m+3=\pm i\sqrt{3}
.\end{array}
Using the properties of equality to isolate the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
m=-3\pm i\sqrt{3}
.\end{array}
Hence, $
m=\left\{ -3-i\sqrt{3},-3+i\sqrt{3} \right\}
.$