-x^2-16+8*x>=0 inequation

Given the inequality:
$$8 x + \left(- x^{2} - 16\right) \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$8 x + \left(- x^{2} - 16\right) = 0$$
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 - it is the discriminant.
Because
$$a = -1$$
$$b = 8$$
$$c = -16$$
, then

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

(8)^2 - 4 * (-1) * (-16) = 0

Because D = 0, then the equation has one root.

x = -b/2a = -8/2/(-1)

$$x_{1} = 4$$
$$x_{1} = 4$$
$$x_{1} = 4$$
This roots
$$x_{1} = 4$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 4$$
=
$$\frac{39}{10}$$
substitute to the expression
$$8 x + \left(- x^{2} - 16\right) \geq 0$$
$$\left(-16 - \left(\frac{39}{10}\right)^{2}\right) + \frac{8 \cdot 39}{10} \geq 0$$

-1/100 >= 0

but

-1/100 < 0

Then
$$x \leq 4$$
no execute
the solution of our inequality is:
$$x \geq 4$$

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