Answer
$x=\left\{ \dfrac{-5-i\sqrt{55}}{8},\dfrac{-5+i\sqrt{55}}{8} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the non real complex solutions of the given equation, $
4x^2+5x+5=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{4x^2+5x+5}{4}=\dfrac{0}{4}
\\\\
x^2+\dfrac{5}{4}x+\dfrac{5}{4}=0
\\\\
x^2+\dfrac{5}{4}x=-\dfrac{5}{4}
.\end{array}
In the equation above, $b=
\dfrac{5}{4}
.$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then
\begin{array}{l}\require{cancel}
\left( \dfrac{\dfrac{5}{4}}{2} \right)^2
\\\\=
\left( \dfrac{5}{4}\div2 \right)^2
\\\\=
\left( \dfrac{5}{4}\cdot\dfrac{1}{2} \right)^2
\\\\=
\left( \dfrac{5}{8} \right)^2
\\\\=
\dfrac{25}{64}
.\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+\dfrac{5}{4}x+\dfrac{25}{64}=-\dfrac{5}{4}+\dfrac{25}{64}
\\\\
\left( x+\dfrac{5}{8} \right)^2=-\dfrac{80}{64}+\dfrac{25}{64}
\\\\
\left( x+\dfrac{5}{8} \right)^2=-\dfrac{55}{64}
.\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}{8}=\pm\sqrt{-\dfrac{55}{64}}
.\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}{8}=\pm\sqrt{-1}\cdot\sqrt{\dfrac{55}{64}}
.\end{array}
Using $i=\sqrt{-1},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x+\dfrac{5}{8}=\pm i\sqrt{\dfrac{55}{64}}
.\end{array}
Writing the radicand as an expression that contains a factor that is a perfect power of the given index and then extracting the root of that factor, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x+\dfrac{5}{8}=\pm i\sqrt{\dfrac{1}{64}\cdot55}
\\\\
x+\dfrac{5}{8}=\pm i\sqrt{\left(\dfrac{1}{8}\right)^2\cdot55}
\\\\
x+\dfrac{5}{8}=\pm i\left(\dfrac{1}{8}\right)\sqrt{55}
\\\\
x+\dfrac{5}{8}=\pm\dfrac{i\sqrt{55}}{8}
.\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}{8}\pm\dfrac{i\sqrt{55}}{8}
\\\\
x=\dfrac{-5\pm i\sqrt{55}}{8}
.\end{array}
Hence, $
x=\left\{ \dfrac{-5-i\sqrt{55}}{8},\dfrac{-5+i\sqrt{55}}{8} \right\}
.$
