81x^2>=16 inequation

Given the inequality:
$$81 x^{2} \geq 16$$
To solve this inequality, we must first solve the corresponding equation:
$$81 x^{2} = 16$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$81 x^{2} = 16$$
to
$$81 x^{2} - 16 = 0$$
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
$$a = 81$$
$$b = 0$$
$$c = -16$$
, then

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

(0)^2 - 4 * (81) * (-16) = 5184

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

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

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

or
$$x_{1} = \frac{4}{9}$$
$$x_{2} = - \frac{4}{9}$$
$$x_{1} = \frac{4}{9}$$
$$x_{2} = - \frac{4}{9}$$
$$x_{1} = \frac{4}{9}$$
$$x_{2} = - \frac{4}{9}$$
This roots
$$x_{2} = - \frac{4}{9}$$
$$x_{1} = \frac{4}{9}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{4}{9} + - \frac{1}{10}$$
=
$$- \frac{49}{90}$$
substitute to the expression
$$81 x^{2} \geq 16$$
$$81 \left(- \frac{49}{90}\right)^{2} \geq 16$$

2401      
---- >= 16
100       

one of the solutions of our inequality is:
$$x \leq - \frac{4}{9}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \frac{4}{9}$$
$$x \geq \frac{4}{9}$$