Answer
$x=\left\{ \dfrac{-5-i\sqrt{15}}{2},\dfrac{-5+i\sqrt{15}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the non real complex solutions of the given equation, $
-x^2-5x-10=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{-x^2-5x-10}{-1}=\dfrac{0}{-1}
\\\\
x^2+5x+10=0
\\\\
x^2+5x=-10
.\end{array}
In the equation above, $b=
5
.$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then
\begin{array}{l}\require{cancel}
\left( \dfrac{5}{2} \right)^2
\\\\=
\dfrac{25}{4}
.\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}
x^2+5x+\dfrac{25}{4}=-10+\dfrac{25}{4}
\\\\
\left( x+\dfrac{5}{2} \right)^2=-\dfrac{40}{4}+\dfrac{25}{4}
\\\\
\left( x+\dfrac{5}{2} \right)^2=-\dfrac{15}{4}
.\end{array}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{array}{l}\require{cancel}
x+\dfrac{5}{2}=\pm\sqrt{-\dfrac{15}{4}}
.\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}
x+\dfrac{5}{2}=\pm\sqrt{-1}\cdot\sqrt{\dfrac{15}{4}}
.\end{array}
Using $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x+\dfrac{5}{2}=\pm i\sqrt{\dfrac{15}{4}}
.\end{array}
Writing the radicand as an expression that contains a factor that is a perfect factor of the index and extracting its root, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x+\dfrac{5}{2}=\pm i\sqrt{\dfrac{1}{4}\cdot15}
\\\\
x+\dfrac{5}{2}=\pm i\sqrt{\left(\dfrac{1}{2}\right)^2\cdot15}
\\\\
x+\dfrac{5}{2}=\pm i\left(\dfrac{1}{2}\right)\sqrt{15}
\\\\
x+\dfrac{5}{2}=\pm \dfrac{i\sqrt{15}}{2}
.\end{array}
Using the properties of equality to isolate the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=-\dfrac{5}{2}\pm \dfrac{i\sqrt{15}}{2}
\\\\
x=\dfrac{-5\pm i\sqrt{15}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{-5-i\sqrt{15}}{2},\dfrac{-5+i\sqrt{15}}{2} \right\}
.$