x^2-7x-8>=0 inequation

Given the inequality:
$$x^{2} - 7 x - 8 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$x^{2} - 7 x - 8 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -7$$
$$c = -8$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 1 \cdot 4 \left(-8\right) + \left(-7\right)^{2} = 81$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 8$$
Simplify
$$x_{2} = -1$$
Simplify
$$x_{1} = 8$$
$$x_{2} = -1$$
$$x_{1} = 8$$
$$x_{2} = -1$$
This roots
$$x_{2} = -1$$
$$x_{1} = 8$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-1 - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$x^{2} - 7 x - 8 \geq 0$$
$$\left(-1\right) 8 + \left(- \frac{11}{10}\right)^{2} - 7 \left(- \frac{11}{10}\right) \geq 0$$

 91     
--- >= 0
100     

one of the solutions of our inequality is:
$$x \leq -1$$

 _____           _____          
      \         /
-------•-------•-------
       x_2      x_1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq -1$$
$$x \geq 8$$