-x^2-6x-9>=0 inequation

Given the inequality:
$$- x^{2} - 6 x - 9 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- x^{2} - 6 x - 9 = 0$$
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
$$a = -1$$
$$b = -6$$
$$c = -9$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) \left(\left(-1\right) 4\right) \left(-9\right) + \left(-6\right)^{2} = 0$$
Because D = 0, then the equation has one root.

x = -b/2a = --6/2/(-1)

$$x_{1} = -3$$
$$x_{1} = -3$$
$$x_{1} = -3$$
This roots
$$x_{1} = -3$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-3 - \frac{1}{10}$$
=
$$- \frac{31}{10}$$
substitute to the expression
$$- x^{2} - 6 x - 9 \geq 0$$
$$- \left(- \frac{31}{10}\right)^{2} - 9 - 6 \left(- \frac{31}{10}\right) \geq 0$$

-1/100 >= 0

but

-1/100 < 0

Then
$$x \leq -3$$
no execute
the solution of our inequality is:
$$x \geq -3$$

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