x^2-5x-7>-1 inequation

Given the inequality:
$$\left(x^{2} - 5 x\right) - 7 > -1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(x^{2} - 5 x\right) - 7 = -1$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(x^{2} - 5 x\right) - 7 = -1$$
to
$$\left(\left(x^{2} - 5 x\right) - 7\right) + 1 = 0$$
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
$$a = 1$$
$$b = -5$$
$$c = -6$$
, then

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

(-5)^2 - 4 * (1) * (-6) = 49

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

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

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

or
$$x_{1} = 6$$
$$x_{2} = -1$$
$$x_{1} = 6$$
$$x_{2} = -1$$
$$x_{1} = 6$$
$$x_{2} = -1$$
This roots
$$x_{2} = -1$$
$$x_{1} = 6$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
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
$$\left(x^{2} - 5 x\right) - 7 > -1$$
$$-7 + \left(\left(- \frac{11}{10}\right)^{2} - \frac{\left(-11\right) 5}{10}\right) > -1$$

-29      
---- > -1
100      

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

 _____           _____          
      \         /
-------ο-------ο-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -1$$
$$x > 6$$