Answer
${{x}^{2}}-10x+29=0$
Work Step by Step
$5-2i\ \text{and }5+2i$
It can be expressed as:
$x=5+2i$ and $x=5-2i$
$\begin{align}
& x-\left( 5+2i \right)=0 \\
& \left( x-5 \right)-2i=0
\end{align}$
Or,
$\begin{align}
& x-\left( 5-2i \right)=0 \\
& \left( x-5 \right)+2i=0 \\
\end{align}$
Apply the zero-product property
$\left( \left( x-5 \right)-2i \right)\left( \left( x-5 \right)+2i \right)=0$
$\begin{align}
& \left( \left( x-5 \right)-2i \right)\left( \left( x-5 \right)+2i \right)=0 \\
& \left( x-5 \right)\left( x-5 \right)+2i\left( x-5 \right)-2i\left( x-5 \right)-4{{i}^{2}}=
\end{align}$
Combine like terms,
Here, ${{i}^{2}}=1$ which is a complex entry.
${{\left( x-5 \right)}^{2}}-4\left( {{i}^{2}} \right)=0$
Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{align}
& {{x}^{2}}+25-10x-4\left( -1 \right)=0 \\
& {{x}^{2}}+25-10x+4=0 \\
& {{x}^{2}}-10x+29=0
\end{align}$
