absolute(2x^2-9x+15)>20 inequation

Given the inequality:
$$\left|{\left(2 x^{2} - 9 x\right) + 15}\right| > 20$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{\left(2 x^{2} - 9 x\right) + 15}\right| = 20$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$2 x^{2} - 9 x + 15 \geq 0$$
or
$$-\infty < x \wedge x < \infty$$
we get the equation
$$\left(2 x^{2} - 9 x + 15\right) - 20 = 0$$
after simplifying we get
$$2 x^{2} - 9 x - 5 = 0$$
the solution in this interval:
$$x_{1} = - \frac{1}{2}$$
$$x_{2} = 5$$

2.
$$2 x^{2} - 9 x + 15 < 0$$
The inequality system has no solutions, see the next condition


$$x_{1} = - \frac{1}{2}$$
$$x_{2} = 5$$
$$x_{1} = - \frac{1}{2}$$
$$x_{2} = 5$$
This roots
$$x_{1} = - \frac{1}{2}$$
$$x_{2} = 5$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{2} + - \frac{1}{10}$$
=
$$- \frac{3}{5}$$
substitute to the expression
$$\left|{\left(2 x^{2} - 9 x\right) + 15}\right| > 20$$
$$\left|{\left(2 \left(- \frac{3}{5}\right)^{2} - \frac{\left(-3\right) 9}{5}\right) + 15}\right| > 20$$

528     
--- > 20
 25     

one of the solutions of our inequality is:
$$x < - \frac{1}{2}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{1}{2}$$
$$x > 5$$