How do you solve $25a=10{{a}^{2}}$?

Hint: The equation given to us is a quadratic equation in the variable $a$. So the solution of the given equation will be the two values of $a$. For solving, we first need to write it in the standard form by subtracting $25a$ from both the sides of the equation to obtain it as $10{{a}^{2}}-25a=0$. Then we have to factorize the polynomial on the LHS of the equation by taking the common factors outside. Finally, using the zero product rule, we will get the final solutions.

Complete step by step solution:
The given equation is
\[\Rightarrow 25a=10{{a}^{2}}\]
Subtracting $25a$ from both sides, we get
$\begin{align}
  & \Rightarrow 25a-25a=10{{a}^{2}}-25a \\
 & \Rightarrow 0=10{{a}^{2}}-25a \\
 & \Rightarrow 10{{a}^{2}}-25a=0 \\
\end{align}$
Since $a$ is common to both the terms of the polynomial on the LHS, we take it outside to get
$\Rightarrow a\left( 10a-25 \right)=0$
Now, from the zero product rule we can say
$\Rightarrow a=0$
And
$\Rightarrow 10a-25=0$
Adding $25$ both the sides, we get
$\begin{align}
  & \Rightarrow 10a-25+25=0+25 \\
 & \Rightarrow 10a=25 \\
\end{align}$
Finally, dividing by $10$ we get
$\begin{align}
  & \Rightarrow \dfrac{10a}{10}=\dfrac{25}{10} \\
 & \Rightarrow a=2.5 \\
\end{align}$
Hence, the solutions of the given equation are $a=0$ and $a=2.5$.

Note: We can also solve the given equation using the quadratic formula, which is given as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. For that, we have to substitute the values of the coefficients from the equation $10{{a}^{2}}-25a=0$ which are noted as $a=10$ $b=-25$ and $c=0$. But we must note that the use of this formula involves more calculations and the factoring method is easier. Also, since the given quadratic equation is a special case when $c=0$, the factoring of the polynomial is even more convenient.