How do you solve $15{{x}^{3}}-7{{x}^{2}}-2x=0$ ?
(a) Using linear formulas
(b) Using trigonometric identities
(c) Using algebraic properties
(d) None of these

Hint: We are to find the solution of the given term we will start by taking x common from all the terms. Then, going ahead we will try a middle term process to get the solution. So, we have the solutions by using the linear formulas.

Complete step by step solution:
According to the question, we are to find the solution of the given term, $15{{x}^{3}}-7{{x}^{2}}-2x=0$.
To start with, we can start by taking x common from all the terms given on the left side.
We have, $15{{x}^{3}}-7{{x}^{2}}-2x=0$
Now, taking common, $x\left( 15{{x}^{2}}-7x-2 \right)=0$
For, $x\ne 0$ , we can say, $15{{x}^{2}}-7x-2=0$.
And there can be another case, $x=0$ .
So, now we have,
$15{{x}^{2}}-7x-2=0$
Now, we will try to solve the equation with the middle term process.
We will do that by multiplying the first and last coefficient and then factorize it. We will also try to factoring terms to add and subtract to get the middle term.
So, multiplying the terms, we have, $15\times 2=30$
Now, this can be written as, $30=10\times 3$ so that by subtracting 10 – 3 we get 7.
Writing it, $15{{x}^{2}}-\left( 10-3 \right)x-2=0$
Simplifying, $15{{x}^{2}}-10x+3x-2=0$
Now, again, taking 5x common from the first two terms and 1 from the last two we get,
$\Rightarrow 5x\left( 3x-2 \right)+1\left( 3x-2 \right)=0$
Writing them altogether,
$\Rightarrow \left( 5x+1 \right)\left( 3x-2 \right)=0$
Now, we can write, 5x + 1 = 0 as 5x = -1
Thus, we have, $x=-\dfrac{1}{5}$
And, 3x – 2 = 0 can be written as, 3x = 2
So, $x=\dfrac{2}{3}$
Thus, we get the value of x as, $x=0,\dfrac{2}{3},-\dfrac{1}{5}$

So, the correct answer is “Option A”.

Note: In this problem, we have used the terms to find three solutions of the equation. As the equation is a third degree equation, we have three solutions. The degree of the equation gives us the number of solutions of the equation.