-x²+6x+7≥0 inequation

Given the inequality:
$$\left(- x^{2} + 6 x\right) + 7 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- x^{2} + 6 x\right) + 7 = 0$$
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
$$a = -1$$
$$b = 6$$
$$c = 7$$
, then

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

(6)^2 - 4 * (-1) * (7) = 64

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

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

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

or
$$x_{1} = -1$$
$$x_{2} = 7$$
$$x_{1} = -1$$
$$x_{2} = 7$$
$$x_{1} = -1$$
$$x_{2} = 7$$
This roots
$$x_{1} = -1$$
$$x_{2} = 7$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\left(- x^{2} + 6 x\right) + 7 \geq 0$$
$$\left(\frac{\left(-11\right) 6}{10} - \left(- \frac{11}{10}\right)^{2}\right) + 7 \geq 0$$

-81      
---- >= 0
100      

but

-81     
---- < 0
100     

Then
$$x \leq -1$$
no execute
one of the solutions of our inequality is:
$$x \geq -1 \wedge x \leq 7$$

         _____  
        /     \  
-------•-------•-------
       x1      x2