3^(2x+1)-8*3^(x+1)+45<=0 inequation

Given the inequality:
$$\left(- 8 \cdot 3^{x + 1} + 3^{2 x + 1}\right) + 45 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 8 \cdot 3^{x + 1} + 3^{2 x + 1}\right) + 45 = 0$$
Solve:
Given the equation:
$$\left(- 8 \cdot 3^{x + 1} + 3^{2 x + 1}\right) + 45 = 0$$
or
$$\left(- 8 \cdot 3^{x + 1} + 3^{2 x + 1}\right) + 45 = 0$$
Do replacement
$$v = 3^{x}$$
we get
$$3 v^{2} - 24 v + 45 = 0$$
or
$$3 v^{2} - 24 v + 45 = 0$$
This equation is of the form

a*v^2 + b*v + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 3$$
$$b = -24$$
$$c = 45$$
, then

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

(-24)^2 - 4 * (3) * (45) = 36

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

v1 = (-b + sqrt(D)) / (2*a)

v2 = (-b - sqrt(D)) / (2*a)

or
$$v_{1} = 5$$
$$v_{2} = 3$$
do backward replacement
$$3^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(3 \right)}}$$
$$x_{1} = 3$$
$$x_{2} = 5$$
$$x_{1} = 3$$
$$x_{2} = 5$$
This roots
$$x_{1} = 3$$
$$x_{2} = 5$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
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
$$\left(- 8 \cdot 3^{x + 1} + 3^{2 x + 1}\right) + 45 \leq 0$$
$$45 + \left(- 8 \cdot 3^{1 + \frac{29}{10}} + 3^{1 + \frac{2 \cdot 29}{10}}\right) \leq 0$$

          9/10        4/5     
45 - 216*3     + 729*3    <= 0
     

but

          9/10        4/5     
45 - 216*3     + 729*3    >= 0
     

Then
$$x \leq 3$$
no execute
one of the solutions of our inequality is:
$$x \geq 3 \wedge x \leq 5$$

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