Answer
${{x}^{2}}-4x+53=0$
Work Step by Step
$2-7i\ \text{and }2+7i$
It can be expressed as:
$x=\text{ }2+7i$ and $x=\text{ }2+7i$
$\begin{align}
& x-\left( \text{ }2+7i \right)=0 \\
& \left( x-2 \right)-7i=0
\end{align}$
Or,
$\begin{align}
& x-\left( 2-7i \right)=0 \\
& \left( x-2 \right)+7i=0 \\
\end{align}$
Apply the zero-product property
$\left( \left( x-2 \right)-7i \right)\left( \left( x-2 \right)+7i \right)=0$
$\begin{align}
& \left( \left( x-2 \right)-7i \right)\left( \left( x-2 \right)+7i \right)=0 \\
& \left( x-2 \right)\left( x-2 \right)+7i\left( x-2 \right)-7i\left( x-2 \right)-49{{i}^{2}}=
\end{align}$
Combine like terms,
Here ${{i}^{2}}=1$ which is a complex entry.
${{\left( x-2 \right)}^{2}}-49\left( {{i}^{2}} \right)=0$
Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{align}
& {{x}^{2}}+4-4x-49\left( -1 \right)=0 \\
& {{x}^{2}}+4-4x+49=0 \\
& {{x}^{2}}-4x+53=0
\end{align}$