16*x^2<47 inequation

Given the inequality:
$$16 x^{2} < 47$$
To solve this inequality, we must first solve the corresponding equation:
$$16 x^{2} = 47$$
Solve:
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$16 x^{2} = 47$$
to
$$16 x^{2} - 47 = 0$$
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
$$a = 16$$
$$b = 0$$
$$c = -47$$
, then

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

(0)^2 - 4 * (16) * (-47) = 3008

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

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

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

or
$$x_{1} = \frac{\sqrt{47}}{4}$$
$$x_{2} = - \frac{\sqrt{47}}{4}$$
$$x_{1} = \frac{\sqrt{47}}{4}$$
$$x_{2} = - \frac{\sqrt{47}}{4}$$
$$x_{1} = \frac{\sqrt{47}}{4}$$
$$x_{2} = - \frac{\sqrt{47}}{4}$$
This roots
$$x_{2} = - \frac{\sqrt{47}}{4}$$
$$x_{1} = \frac{\sqrt{47}}{4}$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{\sqrt{47}}{4} - \frac{1}{10}$$
=
$$- \frac{\sqrt{47}}{4} - \frac{1}{10}$$
substitute to the expression
$$16 x^{2} < 47$$
$$16 \left(- \frac{\sqrt{47}}{4} - \frac{1}{10}\right)^{2} < 47$$

                  2     
   /         ____\      
   |  1    \/ 47 |  < 47
16*|- -- - ------|      
   \  10     4   /      

but

                  2     
   /         ____\      
   |  1    \/ 47 |  > 47
16*|- -- - ------|      
   \  10     4   /      

Then
$$x < - \frac{\sqrt{47}}{4}$$
no execute
one of the solutions of our inequality is:
$$x > - \frac{\sqrt{47}}{4} \wedge x < \frac{\sqrt{47}}{4}$$

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