log5-x(x+1)<1 inequation

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

The equation is transformed from
$$- x \left(x + 1\right) + \log{\left(5 \right)} = 1$$
to
$$\left(- x \left(x + 1\right) + \log{\left(5 \right)}\right) - 1 = 0$$
Expand the expression in the equation
$$\left(- x \left(x + 1\right) + \log{\left(5 \right)}\right) - 1 = 0$$
We get the quadratic equation
$$- x^{2} - x - 1 + \log{\left(5 \right)} = 0$$
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
$$a = -1$$
$$b = -1$$
$$c = -1 + \log{\left(5 \right)}$$
, then

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

(-1)^2 - 4 * (-1) * (-1 + log(5)) = -3 + 4*log(5)

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

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

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

or
$$x_{1} = - \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{1}{2}$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2}$$
$$x_{1} = - \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{1}{2}$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2}$$
$$x_{1} = - \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{1}{2}$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2}$$
This roots
$$x_{1} = - \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{1}{2}$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(- \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{1}{2}\right) + - \frac{1}{10}$$
=
$$- \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{3}{5}$$
substitute to the expression
$$- x \left(x + 1\right) + \log{\left(5 \right)} < 1$$
$$- \left(- \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{3}{5}\right) \left(\left(- \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{3}{5}\right) + 1\right) + \log{\left(5 \right)} < 1$$

  /        _______________\ /      _______________\             
  |  3   \/ -3 + 4*log(5) | |2   \/ -3 + 4*log(5) |             
- |- - - -----------------|*|- - -----------------| + log(5) < 1
  \  5           2        / \5           2        /             
    

one of the solutions of our inequality is:
$$x < - \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{1}{2}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2} - \frac{1}{2}$$
$$x > - \frac{1}{2} + \frac{\sqrt{-3 + 4 \log{\left(5 \right)}}}{2}$$