Answer
${{x}^{2}}-6x-5=0$
Work Step by Step
$3-\sqrt{14}\ \text{and }3+\sqrt{14}$
It can be expressed as,
$x=3+\sqrt{14}$ and $x=3-\sqrt{14}$
$\begin{align}
& x-\left( 3+\sqrt{14} \right)=0 \\
& \left( x-3 \right)-\sqrt{14}=0
\end{align}$
Or,
$\begin{align}
& x-\left( 3-\sqrt{14} \right)=0 \\
& \left( x-3 \right)+\sqrt{14}=0 \\
\end{align}$
Apply the zero-product property
$\left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0$
$\begin{align}
& \left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0 \\
& \left( x-3 \right)\left( x-3 \right)+\sqrt{14}\left( x-3 \right)-\sqrt{14}\left( x-3 \right)-{{\left( \sqrt{14} \right)}^{2}}=0
\end{align}$
Combine like terms:
Here ${{i}^{2}}=1$ which is a complex entry.
${{\left( x-3 \right)}^{2}}-{{\left( \sqrt{14} \right)}^{2}}=0$
Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{align}
& {{x}^{2}}+9-6x-\left( 14 \right)=0 \\
& {{x}^{2}}-6x+9-14=0 \\
& {{x}^{2}}-6x-5=0
\end{align}$
