12x−x^2≥0. inequation

Given the inequality:
$$- x^{2} + 12 x \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- x^{2} + 12 x = 0$$
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
$$a = -1$$
$$b = 12$$
$$c = 0$$
, then

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

(12)^2 - 4 * (-1) * (0) = 144

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

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

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

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

-121      
----- >= 0
 100      

but

-121     
----- < 0
 100     

Then
$$x \leq 0$$
no execute
one of the solutions of our inequality is:
$$x \geq 0 \wedge x \leq 12$$

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        /     \  
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       x1      x2