Answer
The solution set of equation $x\left( x-2 \right)=4$ is $\left\{ \left. 1-\sqrt{5},1+\sqrt{5} \right\} \right.$.
Work Step by Step
Consider the equation: $x\left( x-2 \right)=4$
$\begin{align}
& x\left( x-2 \right)=4 \\
& {{x}^{2}}-2x=4 \\
& {{x}^{2}}-2x-4=0 \\
\end{align}$
The above equation is in the form of $a{{x}^{2}}+bx+c=0$
Apply quadratic formula.
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a is 1, b is $-2$ and c is $-4$
$\begin{align}
& x=\frac{2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -4 \right)}}{2\left( 1 \right)} \\
& =\frac{2\pm \sqrt{4+16}}{2} \\
& =\frac{2\pm \sqrt{20}}{2} \\
& =\frac{2\pm \sqrt{5\cdot 4}}{2}
\end{align}$
On further simplification, we get:
$x=\frac{2\pm 2\sqrt{5}}{2}$
Take out the 2 common term:
$\begin{align}
& x=\frac{2\left( 1\pm \sqrt{5} \right)}{2} \\
& =1\pm \sqrt{5}
\end{align}$
Check:
For $x=1+\sqrt{5}$
Put $x=1+\sqrt{5}$ in $x\left( x-2 \right)=4$
$\begin{align}
& \left( 1+\sqrt{5} \right)\left( 1+\sqrt{5}-2 \right)\overset{?}{\mathop{=}}\,4 \\
& \left( \sqrt{5}+1 \right)\left( \sqrt{5}-1 \right)\overset{?}{\mathop{=}}\,4 \\
& {{\left( \sqrt{5} \right)}^{2}}-{{1}^{2}}\overset{?}{\mathop{=}}\,4 \\
& 5-1\overset{?}{\mathop{=}}\,4
\end{align}$
Which is true.
For $x=1-\sqrt{5}$
Put $x=1-\sqrt{5}$ in $x\left( x-2 \right)=4$
$\begin{align}
& \left( 1-\sqrt{5} \right)\left( 1-\sqrt{5}-2 \right)\overset{?}{\mathop{=}}\,4 \\
& \left( 1-\sqrt{5} \right)\left( -1-\sqrt{5} \right)\overset{?}{\mathop{=}}\,4 \\
& {{\left( -\sqrt{5} \right)}^{2}}-{{1}^{2}}\overset{?}{\mathop{=}}\,4 \\
& 5-1\overset{?}{\mathop{=}}\,4
\end{align}$
Which is true.
Therefore, the solution set of the equation $x\left( x-2 \right)=4$ is $\left\{ \left. 1-\sqrt{5},1+\sqrt{5} \right\} \right.$.
