The equation 2x^3 + 9x^2 -18x = 0 has to be solved.

2x^3 + 9x^2 -18x = 0

=> x*(2x^2 + 9x - 18) = 0

=> x*(2x^2 + 12x - 3x - 18) = 0

=> x*(2x(x + 6) - 3(x + 6)) = 0

=> x*(2x-3)(x + 6) = 0

This gives x = 0

2x - 3 = 0

=> x = 3/2

x + 6 = 0

=> x = -6

**The solutions of the equation are {-6, 0, 3/2}**

First, you want to factor out anything that is common in each of the terms. In this equation, that is the x. So:

x(2x^2+9x-18)=0

Then, you separate this into two equations,

x=0 and 2x^2+9x-18=0 (because if x=0, then the whole equation will be equal to zero).

Now you need to find the remaining values of x:

1. Factor: (2x-3)(x+6)=0

2. Separate: 2x-3=0, x+6=0

3. Solve for x in both equations:

2x-3=0

2x=3

x=3/2

and

x+6=0

x=-6

Therefore, the solutions to this equation (called the "zeros" of the function because they occur when the graph crosses the x-axis) are: x=0, x=3/2, and x=6