2x⁴-5x²-12<0 inequation

Given the inequality:
$$2 x^{4} - 5 x^{2} - 12 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 x^{4} - 5 x^{2} - 12 = 0$$
Solve:
Given the equation:
$$2 x^{4} - 5 x^{2} - 12 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$2 v^{2} - 5 v - 12 = 0$$
This equation is of the form

a*v^2 + b*v + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 2$$
$$b = -5$$
$$c = -12$$
, then

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

(-5)^2 - 4 * (2) * (-12) = 121

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

v1 = (-b + sqrt(D)) / (2*a)

v2 = (-b - sqrt(D)) / (2*a)

or
$$v_{1} = 4$$
Simplify
$$v_{2} = - \frac{3}{2}$$
Simplify
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = 4$$
$$x_{2} = - \frac{3}{2}$$
$$x_{1} = 4$$
$$x_{2} = - \frac{3}{2}$$
This roots
$$x_{2} = - \frac{3}{2}$$
$$x_{1} = 4$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{3}{2} - \frac{1}{10}$$
=
$$- \frac{8}{5}$$
substitute to the expression
$$2 x^{4} - 5 x^{2} - 12 < 0$$
$$- 5 \left(- \frac{8}{5}\right)^{2} - 12 + 2 \left(- \frac{8}{5}\right)^{4} < 0$$

-7308     
------ < 0
 625      

one of the solutions of our inequality is:
$$x < - \frac{3}{2}$$

 _____           _____          
      \         /
-------ο-------ο-------
       x_2      x_1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{3}{2}$$
$$x > 4$$