Answer
The solution of the equation is $\underline{x=\left\{ 3,4 \right\}}$
Work Step by Step
Let us consider the equation, $\sqrt{{{x}^{2}}-3x}=2x-6$
By squaring both sides, we obtain:
$\begin{align}
& {{\left( \sqrt{{{x}^{2}}-3x} \right)}^{2}}={{\left( 2x-6 \right)}^{2}} \\
& {{x}^{2}}-3x={{\left( 2x \right)}^{2}}-2\left( 2x \right)\left( -6 \right)+{{\left( 6 \right)}^{2}} \\
& {{x}^{2}}-3x=4{{x}^{2}}-24x+36 \\
& 3{{x}^{2}}-21x+36=0
\end{align}$
And rearrange the equation as given below:
$\begin{align}
& {{x}^{2}}-7x+12=0 \\
& {{x}^{2}}-3x-4x+12=0 \\
& x\left( x-3 \right)-4\left( x-3 \right)=0 \\
& \left( x-3 \right)\left( x-4 \right)=0 \\
& x=3,4
\end{align}$
Thus, the value of $ x=3,4$.
