-3x^2+16x-5<0 inequation

Given the inequality:
$$\left(- 3 x^{2} + 16 x\right) - 5 < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 3 x^{2} + 16 x\right) - 5 = 0$$
Solve:
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
$$a = -3$$
$$b = 16$$
$$c = -5$$
, then

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

(16)^2 - 4 * (-3) * (-5) = 196

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

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

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

or
$$x_{1} = \frac{1}{3}$$
$$x_{2} = 5$$
$$x_{1} = \frac{1}{3}$$
$$x_{2} = 5$$
$$x_{1} = \frac{1}{3}$$
$$x_{2} = 5$$
This roots
$$x_{1} = \frac{1}{3}$$
$$x_{2} = 5$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{3}$$
=
$$\frac{7}{30}$$
substitute to the expression
$$\left(- 3 x^{2} + 16 x\right) - 5 < 0$$
$$-5 + \left(- 3 \left(\frac{7}{30}\right)^{2} + \frac{7 \cdot 16}{30}\right) < 0$$

-143     
----- < 0
 100     

one of the solutions of our inequality is:
$$x < \frac{1}{3}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{1}{3}$$
$$x > 5$$