x^2+x-12=>0 inequation

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

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

(1)^2 - 4 * (1) * (-12) = 49

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

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

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

or
$$x_{1} = 3$$
$$x_{2} = -4$$
$$x_{1} = 3$$
$$x_{2} = -4$$
$$x_{1} = 3$$
$$x_{2} = -4$$
This roots
$$x_{2} = -4$$
$$x_{1} = 3$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-4 + - \frac{1}{10}$$
=
$$- \frac{41}{10}$$
substitute to the expression
$$\left(x^{2} + x\right) - 12 \geq 0$$
$$-12 + \left(- \frac{41}{10} + \left(- \frac{41}{10}\right)^{2}\right) \geq 0$$

 71     
--- >= 0
100     

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

 _____           _____          
      \         /
-------•-------•-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq -4$$
$$x \geq 3$$