3x²+2x-5 inequation

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

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

(2)^2 - 4 * (3) * (-5) = 64

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

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

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

or
$$x_{1} = 1$$
$$x_{2} = - \frac{5}{3}$$
$$x_{1} = 1$$
$$x_{2} = - \frac{5}{3}$$
$$x_{1} = 1$$
$$x_{2} = - \frac{5}{3}$$
This roots
$$x_{2} = - \frac{5}{3}$$
$$x_{1} = 1$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{5}{3} + - \frac{1}{10}$$
=
$$- \frac{53}{30}$$
substitute to the expression
$$\left(3 x^{2} + 2 x\right) - 5 > 0$$
$$-5 + \left(\frac{\left(-53\right) 2}{30} + 3 \left(- \frac{53}{30}\right)^{2}\right) > 0$$

 83    
--- > 0
100    

one of the solutions of our inequality is:
$$x < - \frac{5}{3}$$

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

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