4x^2-9x-9=>0 inequation

Given the inequality:
$$\left(4 x^{2} - 9 x\right) - 9 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(4 x^{2} - 9 x\right) - 9 = 0$$
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
$$a = 4$$
$$b = -9$$
$$c = -9$$
, then

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

(-9)^2 - 4 * (4) * (-9) = 225

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

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

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

or
$$x_{1} = 3$$
$$x_{2} = - \frac{3}{4}$$
$$x_{1} = 3$$
$$x_{2} = - \frac{3}{4}$$
$$x_{1} = 3$$
$$x_{2} = - \frac{3}{4}$$
This roots
$$x_{2} = - \frac{3}{4}$$
$$x_{1} = 3$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{3}{4} + - \frac{1}{10}$$
=
$$- \frac{17}{20}$$
substitute to the expression
$$\left(4 x^{2} - 9 x\right) - 9 \geq 0$$
$$-9 + \left(4 \left(- \frac{17}{20}\right)^{2} - \frac{\left(-17\right) 9}{20}\right) \geq 0$$

77     
-- >= 0
50     

one of the solutions of our inequality is:
$$x \leq - \frac{3}{4}$$

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

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