How do you factor $5{{x}^{2}}-35x$?

Hint: We are given a quadratic equation which has to be solved by the method of factoring the equation. We shall break down the x-variable term into two parts which must add up to 35x given in the equation. however, due to absence of the constant term, we shall directly take common from both the terms and group them. Further, we shall equate both groups equal to zero to find the two roots of the equation.

Complete step by step solution:
There are four methods for solving quadratic equations, namely, factoring method, completing the square method, taking the square root method and the last method is solving using the various properties of polynomials.
However, we will use the method of factoring the quadratic equation which makes our calculations simpler.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $5{{x}^{2}}-35x$,
$\Rightarrow 5{{x}^{2}}-35x=0$
Taking common, we get:
$\begin{align}
  & \Rightarrow 5x\left( x-7 \right)=0 \\
 & \Rightarrow \left( 5x \right)\left( x-7 \right)=0 \\
\end{align}$
Now, we shall equate both the groups equal to zero.
Hence, $5x=0$ or $x-7=0$
For $5x=0$, dividing both sides by 5, we get
$\Rightarrow x=0$
For $x-7=0$, transposing 7 to the right hand side, we get
 $\Rightarrow x=7$
Therefore, the roots of the equation are $x=0,7$.

Note: We must be careful enough to find the right common factors of the multiple terms given. We must also be careful enough while grouping the terms in order to avoid mistakes. One possible mistake that could be made while transposing terms that 7 could have been written as -7 which would produce incorrect roots of the equation.