(z-5)*(z+5)>0 inequation

Given the inequality:
$$\left(z - 5\right) \left(z + 5\right) > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(z - 5\right) \left(z + 5\right) = 0$$
Solve:
Expand the expression in the equation
$$\left(z - 5\right) \left(z + 5\right) = 0$$
We get the quadratic equation
$$z^{2} - 25 = 0$$
This equation is of the form

a*z^2 + b*z + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = -25$$
, then

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

(0)^2 - 4 * (1) * (-25) = 100

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

z1 = (-b + sqrt(D)) / (2*a)

z2 = (-b - sqrt(D)) / (2*a)

or
$$z_{1} = 5$$
$$z_{2} = -5$$
$$z_{1} = 5$$
$$z_{2} = -5$$
$$z_{1} = 5$$
$$z_{2} = -5$$
This roots
$$z_{2} = -5$$
$$z_{1} = 5$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$z_{0} < z_{2}$$
For example, let's take the point
$$z_{0} = z_{2} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$- \frac{51}{10}$$
substitute to the expression
$$\left(z - 5\right) \left(z + 5\right) > 0$$
$$\left(- \frac{51}{10} - 5\right) \left(- \frac{51}{10} + 5\right) > 0$$

101    
--- > 0
100    

one of the solutions of our inequality is:
$$z < -5$$

 _____           _____          
      \         /
-------ο-------ο-------
       z2      z1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$z < -5$$
$$z > 5$$