Answer
${{x}^{2}}-4x-6=0$
Work Step by Step
$2-\sqrt{10}\ \text{and 2}+\sqrt{10}$
It can be expressed as:
$x=\text{2}+\sqrt{10}$ and $x=2-\sqrt{10}$
$\begin{align}
& x-\left( \text{2}+\sqrt{10} \right)=0 \\
& \left( x-2 \right)-\sqrt{10}=0
\end{align}$
Or,
$\begin{align}
& x-\left( 2-\sqrt{10} \right)=0 \\
& \left( x-2 \right)+\sqrt{10}=0 \\
\end{align}$
Apply the zero-product property
$\left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0$
$\begin{align}
& \left( \left( x-2 \right)-\sqrt{10} \right)\left( \left( x-2 \right)+\sqrt{10} \right)=0 \\
& \left( x-2 \right)\left( x-2 \right)+\sqrt{10}\left( x-2 \right)-\sqrt{10}\left( x-2 \right)-{{\left( \sqrt{10} \right)}^{2}}=0
\end{align}$
Combine like terms:
${{\left( x-2 \right)}^{2}}-{{\left( \sqrt{10} \right)}^{2}}=0$
Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{align}
& {{x}^{2}}+4-4x-\left( 10 \right)=0 \\
& {{x}^{2}}-4x+4-10=0 \\
& {{x}^{2}}-4x-6=0
\end{align}$
