(2-5x^2)^2>=16 inequation

Given the inequality:
$$\left(2 - 5 x^{2}\right)^{2} \geq 16$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(2 - 5 x^{2}\right)^{2} = 16$$
Solve:
Given the equation:
$$\left(2 - 5 x^{2}\right)^{2} = 16$$
transform:
Take common factor from the equation
$$\left(5 x^{2} - 6\right) \left(5 x^{2} + 2\right) = 0$$
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$5 x^{2} - 6 = 0$$
$$5 x^{2} + 2 = 0$$
solve the resulting equation:
1.
$$5 x^{2} - 6 = 0$$
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
$$a = 5$$
$$b = 0$$
$$c = -6$$
, then

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

(0)^2 - 4 * (5) * (-6) = 120

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

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

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

or
$$x_{1} = \frac{\sqrt{30}}{5}$$
$$x_{2} = - \frac{\sqrt{30}}{5}$$
2.
$$5 x^{2} + 2 = 0$$
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:
$$x_{3} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{4} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 5$$
$$b = 0$$
$$c = 2$$
, then

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

(0)^2 - 4 * (5) * (2) = -40

Because D<0, then the equation
has no real roots,
but complex roots is exists.

x3 = (-b + sqrt(D)) / (2*a)

x4 = (-b - sqrt(D)) / (2*a)

or
$$x_{3} = \frac{\sqrt{10} i}{5}$$
$$x_{4} = - \frac{\sqrt{10} i}{5}$$
$$x_{1} = \frac{\sqrt{30}}{5}$$
$$x_{2} = - \frac{\sqrt{30}}{5}$$
$$x_{3} = \frac{\sqrt{10} i}{5}$$
$$x_{4} = - \frac{\sqrt{10} i}{5}$$
Exclude the complex solutions:
$$x_{1} = \frac{\sqrt{30}}{5}$$
$$x_{2} = - \frac{\sqrt{30}}{5}$$
This roots
$$x_{2} = - \frac{\sqrt{30}}{5}$$
$$x_{1} = \frac{\sqrt{30}}{5}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{\sqrt{30}}{5} - \frac{1}{10}$$
=
$$- \frac{\sqrt{30}}{5} - \frac{1}{10}$$
substitute to the expression
$$\left(2 - 5 x^{2}\right)^{2} \geq 16$$
$$\left(2 - 5 \left(- \frac{\sqrt{30}}{5} - \frac{1}{10}\right)^{2}\right)^{2} \geq 16$$

                        2      
/                     2\       
|      /         ____\ |       
|      |  1    \/ 30 | |  >= 16
|2 - 5*|- -- - ------| |       
\      \  10     5   / /       
      

one of the solutions of our inequality is:
$$x \leq - \frac{\sqrt{30}}{5}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \frac{\sqrt{30}}{5}$$
$$x \geq \frac{\sqrt{30}}{5}$$