Answer
$16{{x}^{2}}-40x-8=0$
Work Step by Step
$\frac{5}{4}-\frac{\sqrt{33}}{4}\ \text{and }\frac{5}{4}+\frac{\sqrt{33}}{4}$
It can be expressed as:
$x=\frac{5}{4}+\frac{\sqrt{33}}{4}$ and $x=\frac{5}{4}-\frac{\sqrt{33}}{4}$
$\begin{align}
& x-\left( \frac{5}{4}+\frac{\sqrt{33}}{4} \right)=0 \\
& \left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4}=0
\end{align}$
Or,
$\begin{align}
& x-\left( \frac{5}{4}-\frac{\sqrt{33}}{4} \right)=0 \\
& \left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4}=0 \\
\end{align}$
Apply the zero-product property:
$\left( \left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4} \right)\left( \left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4} \right)=0$
$\begin{align}
& \left( \left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4} \right)\left( \left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4} \right)=0 \\
& \left( x-\frac{5}{4} \right)\left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4}\left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4}\left( x-\frac{5}{4} \right)-{{\left( \frac{\sqrt{33}}{4} \right)}^{2}}=0
\end{align}$
Combine the like terms:
${{\left( x-\frac{5}{4} \right)}^{2}}-{{\left( \frac{\sqrt{33}}{4} \right)}^{2}}=0$
Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\begin{align}
& {{\left( x-\frac{5}{4} \right)}^{2}}-{{\left( \frac{\sqrt{33}}{4} \right)}^{2}}=0 \\
& {{x}^{2}}+\frac{25}{16}-2x\cdot \frac{5}{4}-\frac{33}{16}=0 \\
& {{x}^{2}}-\frac{5x}{2}+\frac{25-33}{16}=0 \\
& {{x}^{2}}-\frac{5x}{2}-\frac{8}{16}=0
\end{align}$
Multiply the equation by $16$ to clear the fraction,
$\begin{align}
& 16\left( {{x}^{2}}-\frac{5x}{2}-\frac{8}{16} \right)=0 \\
& 16{{x}^{2}}-40x-8=0
\end{align}$
