sqrt(2x+10)<3x-5 inequation

Given the inequality:
$$\sqrt{2 x + 10} < 3 x - 5$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2 x + 10} = 3 x - 5$$
Solve:
Given the equation
$$\sqrt{2 x + 10} = 3 x - 5$$
$$\sqrt{2 x + 10} = 3 x - 5$$
We raise the equation sides to 2-th degree
$$2 x + 10 = \left(3 x - 5\right)^{2}$$
$$2 x + 10 = 9 x^{2} - 30 x + 25$$
Transfer the right side of the equation left part with negative sign
$$- 9 x^{2} + 32 x - 15 = 0$$
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
$$a = -9$$
$$b = 32$$
$$c = -15$$
, then

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

(32)^2 - 4 * (-9) * (-15) = 484

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

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

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

or
$$x_{1} = \frac{5}{9}$$
Simplify
$$x_{2} = 3$$
Simplify

Because
$$\sqrt{2 x + 10} = 3 x - 5$$
and
$$\sqrt{2 x + 10} \geq 0$$
then
$$3 x - 5 \geq 0$$
or
$$\frac{5}{3} \leq x$$
$$x < \infty$$
$$x_{2} = 3$$
$$x_{1} = 3$$
$$x_{1} = 3$$
This roots
$$x_{1} = 3$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 3$$
=
$$\frac{29}{10}$$
substitute to the expression
$$\sqrt{2 x + 10} < 3 x - 5$$
$$\sqrt{2 \cdot \frac{29}{10} + 10} < \left(-1\right) 5 + 3 \cdot \frac{29}{10}$$

  _____     
\/ 395    37
------- < --
   5      10
     

but

  _____     
\/ 395    37
------- > --
   5      10
     

Then
$$x < 3$$
no execute
the solution of our inequality is:
$$x > 3$$

         _____  
        /
-------ο-------
       x_1