How do you factor $5{{a}^{2}}+10ab-15{{b}^{2}}$?

Hint: We are given a quadratic equation which has to be solved by the method of factoring the equation. We shall break down the ab-variable term into two parts which must add up to 10ab given in the equation. Then, we will group the common terms and form two linear equations in a-variable and b-variable out of the given quadratic equation. After further equating each of the linear equations to zero to get our required solution.

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, $5{{a}^{2}}+10ab-15{{b}^{2}}$,
$\Rightarrow 5{{a}^{2}}+10ab-15{{b}^{2}}=0$
We will find numbers by hit and trial whose product is equal to $5{{a}^{2}}\times \left( -15{{b}^{2}} \right)=-75{{a}^{2}}{{b}^{2}}$ and whose sum is equal to 10ab.
Such two numbers are 15ab and -5ab as $-5ab+15ab=10ab$ and $\left( -5ab \right)\times 15ab=-75{{a}^{2}}{{b}^{2}}$.
Now, factoring the equation:
$\Rightarrow 5{{a}^{2}}-5ab+15ab-15{{b}^{2}}=0$
Taking common, we get:
$\begin{align}
  & \Rightarrow 5a\left( a-b \right)+15b\left( a-b \right)=0 \\
 & \Rightarrow \left( a-b \right)\left( 5a+15b \right)=0 \\
\end{align}$
Hence, we get $a-b=0$
$\Rightarrow a=b$ …………… equation (1)
And $5a+15b=0$ ………….. equation (2)
Now, we shall solve equations (1) and (2) simultaneously to find the values of a and b.
Substituting from equation (1) to equation (2), we get
$\Rightarrow 5\left( b \right)+15b=0$
$\Rightarrow 20b=0$
$\Rightarrow b=\dfrac{0}{20}$
$\Rightarrow b=0$
Putting this value of b in equation (1), we get
$\Rightarrow a=0$
Therefore, in equation $5{{a}^{2}}+10ab-15{{b}^{2}}$, $a=0$ and $b=0$.

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 was while simultaneously solving the two equations, we could have written -b in place of b during substitution.