How do you solve:
$2x\left( x-5 \right)=0$

Hint: To solve the above equation which is given as: $2x\left( x-5 \right)=0$. First of all, we are going to divide 2 on both the sides then 2 will be eliminated by the L.H.S of the equation and on the R.H.S, dividing 0 by something we get 0. On the R.H.S of the equation, we have 0. Now, the expression $x\left( x-5 \right)$ is equal to 0 when either $x$ is equal to 0 or $x-5$ is equal to 0. So, equating $x$ to 0 and $x-5$ to 0 we will get the solutions of x.

Complete step by step answer:
The equation given above which we have to find the solutions of:
$2x\left( x-5 \right)=0$
Now, we are dividing 2 on both the sides we get,
$\dfrac{2x\left( x-5 \right)}{2}=\dfrac{0}{2}$
On the L.H.S of the above equation, 2 will be cancelled out from the numerator and the denominator while on the R.H.S. we will get 0 because dividing 0 with any number except 0 is 0.
$x\left( x-5 \right)=0$
Now, L.H.S of the above equation is 0 when either $x$ is equal to 0 or $x-5$ is equal to 0. Equating $x\And \left( x-5 \right)$ to 0 we get,
$\begin{align}
  & x=0, \\
 & x-5=0 \\
\end{align}$
One of the solutions of $x$, we have got as 0 and to find the other solution, we are going to add 5 on both sides of the equation $x-5=0$.
$\begin{align}
  & x-5=0 \\
 & \Rightarrow x-5+5=0+5 \\
 & \Rightarrow x=5 \\
\end{align}$
From the above, we got the solutions of the equation as:
$\begin{align}
  & x=0; \\
 & x=5 \\
\end{align}$

Note: The solutions that we got above can be checked by substituting each value of x in $2x\left( x-5 \right)=0$ and see whether those solutions hold true or not. Now, checking the first solution $x=0$ by substituting this value in the above equation we get,
\[\begin{align}
  & 2\left( 0 \right)\left( 0-5 \right)=0 \\
 & \Rightarrow -10\left( 0 \right)=0 \\
\end{align}\]
When we multiply 0 by -10 we will get 0 because multiplying 0 by any number we get 0.
$0=0$
In the above, L.H.S is equal to R.H.S so the solution $x=0$ is correct.
Now, checking $x=5$ by substituting this value of x in $2x\left( x-5 \right)=0$ we get,
$\begin{align}
  & 2\left( 5 \right)\left( 5-5 \right)=0 \\
 & \Rightarrow 10\left( 0 \right)=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
In the above, we got L.H.S equal to R.H.S. This means that the solution $x=5$ is correct.