12(x-10)(x+4)>0 inequation

Given the inequality:
$$12 \left(x - 10\right) \left(x + 4\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$12 \left(x - 10\right) \left(x + 4\right) = 0$$
Solve:
Expand the expression in the equation
$$12 \left(x - 10\right) \left(x + 4\right) = 0$$
We get the quadratic equation
$$12 x^{2} - 72 x - 480 = 0$$
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
$$a = 12$$
$$b = -72$$
$$c = -480$$
, then

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

(-72)^2 - 4 * (12) * (-480) = 28224

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

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

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

or
$$x_{1} = 10$$
$$x_{2} = -4$$
$$x_{1} = 10$$
$$x_{2} = -4$$
$$x_{1} = 10$$
$$x_{2} = -4$$
This roots
$$x_{2} = -4$$
$$x_{1} = 10$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-4 + - \frac{1}{10}$$
=
$$- \frac{41}{10}$$
substitute to the expression
$$12 \left(x - 10\right) \left(x + 4\right) > 0$$
$$12 \left(-10 + - \frac{41}{10}\right) \left(- \frac{41}{10} + 4\right) > 0$$

423    
--- > 0
 25    

one of the solutions of our inequality is:
$$x < -4$$

 _____           _____          
      \         /
-------ο-------ο-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -4$$
$$x > 10$$