-16/(x^2+2)-5>0 inequation

Given the inequality:
$$-5 - \frac{16}{x^{2} + 2} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$-5 - \frac{16}{x^{2} + 2} = 0$$
Solve:
Given the equation:
$$-5 - \frac{16}{x^{2} + 2} = 0$$
Multiply the equation sides by the denominators:
2 + x^2
we get:
$$\left(-5 - \frac{16}{x^{2} + 2}\right) \left(x^{2} + 2\right) = 0$$
$$- 5 x^{2} - 26 = 0$$
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
$$a = -5$$
$$b = 0$$
$$c = -26$$
, then

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

(0)^2 - 4 * (-5) * (-26) = -520

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

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

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

or
$$x_{1} = - \frac{\sqrt{130} i}{5}$$
$$x_{2} = \frac{\sqrt{130} i}{5}$$
$$x_{1} = - \frac{\sqrt{130} i}{5}$$
$$x_{2} = \frac{\sqrt{130} i}{5}$$
Exclude the complex solutions:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example

x0 = 0

$$- \frac{16}{0^{2} + 2} - 5 > 0$$

-13 > 0

so the inequality has no solutions