0.5x^2>13 inequation

Given the inequality:
$$\frac{x^{2}}{2} > 13$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x^{2}}{2} = 13$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\frac{x^{2}}{2} = 13$$
to
$$\frac{x^{2}}{2} - 13 = 0$$
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
$$a = \frac{1}{2}$$
$$b = 0$$
$$c = -13$$
, then

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

(0)^2 - 4 * (1/2) * (-13) = 26

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

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

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

or
$$x_{1} = \sqrt{26}$$
$$x_{2} = - \sqrt{26}$$
$$x_{1} = \sqrt{26}$$
$$x_{2} = - \sqrt{26}$$
$$x_{1} = \sqrt{26}$$
$$x_{2} = - \sqrt{26}$$
This roots
$$x_{2} = - \sqrt{26}$$
$$x_{1} = \sqrt{26}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \sqrt{26} - \frac{1}{10}$$
=
$$- \sqrt{26} - \frac{1}{10}$$
substitute to the expression
$$\frac{x^{2}}{2} > 13$$
$$\frac{\left(- \sqrt{26} - \frac{1}{10}\right)^{2}}{2} > 13$$

               2     
/  1      ____\      
|- -- - \/ 26 |      
\  10         /  > 13
----------------     
       2             
     

one of the solutions of our inequality is:
$$x < - \sqrt{26}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \sqrt{26}$$
$$x > \sqrt{26}$$