#### Answer

$x=\left\{ -4-\sqrt{5},-4+\sqrt{5} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
x^2+8x+11=0
,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to 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}
x^2+8x=-11
.\end{array}
In the equation above, $b=
8
.$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to
\begin{array}{l}\require{cancel}
\left( \dfrac{8}{2} \right)^2
\\\\=
\left( 4 \right)^2
\\\\=
16
.\end{array}
Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to
\begin{array}{l}\require{cancel}
x^2+8x+16=-11+16
\\\\
x^2+8x+16=5
.\end{array}
With the left side now a perfect square trinomial, the equation above is equivalent to
\begin{array}{l}\require{cancel}
(x+4)^2=5
.\end{array}
Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x+4=\pm\sqrt{5}
\\\\
x=-4\pm\sqrt{5}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=-4-\sqrt{5}
\\\\\text{OR}\\\\
x=-4+\sqrt{5}
.\end{array}
Hence, $
x=\left\{ -4-\sqrt{5},-4+\sqrt{5} \right\}
.$