8x²+24x>=0 inequation

Given the inequality:
$$8 x^{2} + 24 x \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$8 x^{2} + 24 x = 0$$
Solve:
This equation is of the form

a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 8$$
$$b = 24$$
$$c = 0$$
, then

D = b^2 - 4 * a * c = 

(24)^2 - 4 * (8) * (0) = 576

Because D > 0, then the equation has two roots.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = 0$$
Simplify
$$x_{2} = -3$$
Simplify
$$x_{1} = 0$$
$$x_{2} = -3$$
$$x_{1} = 0$$
$$x_{2} = -3$$
This roots
$$x_{2} = -3$$
$$x_{1} = 0$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-3 - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$8 x^{2} + 24 x \geq 0$$
$$24 \left(- \frac{31}{10}\right) + 8 \left(- \frac{31}{10}\right)^{2} \geq 0$$

62     
-- >= 0
25     

one of the solutions of our inequality is:
$$x \leq -3$$

 _____           _____          
      \         /
-------•-------•-------
       x_2      x_1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq -3$$
$$x \geq 0$$