(x^2+5*x+12)/(x^2-4*x+5)>3 inequation

Given the inequality:
$$\frac{\left(x^{2} + 5 x\right) + 12}{\left(x^{2} - 4 x\right) + 5} > 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(x^{2} + 5 x\right) + 12}{\left(x^{2} - 4 x\right) + 5} = 3$$
Solve:
Given the equation:
$$\frac{\left(x^{2} + 5 x\right) + 12}{\left(x^{2} - 4 x\right) + 5} = 3$$
Multiply the equation sides by the denominators:
5 + x^2 - 4*x
we get:
$$\frac{\left(\left(x^{2} + 5 x\right) + 12\right) \left(x^{2} - 4 x + 5\right)}{\left(x^{2} - 4 x\right) + 5} = 3 x^{2} - 12 x + 15$$
$$x^{2} + 5 x + 12 = 3 x^{2} - 12 x + 15$$
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$x^{2} + 5 x + 12 = 3 x^{2} - 12 x + 15$$
to
$$- 2 x^{2} + 17 x - 3 = 0$$
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
$$a = -2$$
$$b = 17$$
$$c = -3$$
, then

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

(17)^2 - 4 * (-2) * (-3) = 265

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

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

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

or
$$x_{1} = \frac{17}{4} - \frac{\sqrt{265}}{4}$$
$$x_{2} = \frac{\sqrt{265}}{4} + \frac{17}{4}$$
$$x_{1} = \frac{17}{4} - \frac{\sqrt{265}}{4}$$
$$x_{2} = \frac{\sqrt{265}}{4} + \frac{17}{4}$$
$$x_{1} = \frac{17}{4} - \frac{\sqrt{265}}{4}$$
$$x_{2} = \frac{\sqrt{265}}{4} + \frac{17}{4}$$
This roots
$$x_{1} = \frac{17}{4} - \frac{\sqrt{265}}{4}$$
$$x_{2} = \frac{\sqrt{265}}{4} + \frac{17}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \left(\frac{17}{4} - \frac{\sqrt{265}}{4}\right)$$
=
$$\frac{83}{20} - \frac{\sqrt{265}}{4}$$
substitute to the expression
$$\frac{\left(x^{2} + 5 x\right) + 12}{\left(x^{2} - 4 x\right) + 5} > 3$$
$$\frac{\left(\left(\frac{83}{20} - \frac{\sqrt{265}}{4}\right)^{2} + 5 \left(\frac{83}{20} - \frac{\sqrt{265}}{4}\right)\right) + 12}{\left(- 4 \left(\frac{83}{20} - \frac{\sqrt{265}}{4}\right) + \left(\frac{83}{20} - \frac{\sqrt{265}}{4}\right)^{2}\right) + 5} > 3$$

                    2                
      /       _____\        _____    
131   |83   \/ 265 |    5*\/ 265     
--- + |-- - -------|  - ---------    
 4    \20      4   /        4        
--------------------------------- > 3
                                2    
                  /       _____\     
   58     _____   |83   \/ 265 |     
 - -- + \/ 265  + |-- - -------|     
   5              \20      4   /     

Then
$$x < \frac{17}{4} - \frac{\sqrt{265}}{4}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{17}{4} - \frac{\sqrt{265}}{4} \wedge x < \frac{\sqrt{265}}{4} + \frac{17}{4}$$

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