x^2(-17x)+72>0 inequation

Given the inequality:
$$- 17 x x^{2} + 72 > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$- 17 x x^{2} + 72 = 0$$
Solve:
Given the equation
$$- 17 x x^{2} + 72 = 0$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{17} \sqrt[3]{x^{3}} = \sqrt[3]{72}$$
or
$$\sqrt[3]{17} x = 2 \cdot 3^{\frac{2}{3}}$$
Expand brackets in the left part

x*17^1/3 = 2*3^(2/3)

Expand brackets in the right part

x*17^1/3 = 2*3^2/3

Divide both parts of the equation by 17^(1/3)

x = 2*3^(2/3) / (17^(1/3))

We get the answer: x = 2*51^(2/3)/17

All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = \frac{72}{17}$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = \frac{72}{17}$$
where
$$r = \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = 1$$
so
$$\cos{\left(3 p \right)} = 1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$
$$z_{2} = - \frac{51^{\frac{2}{3}}}{17} - \frac{3 \cdot 17^{\frac{2}{3}} \sqrt[6]{3} i}{17}$$
$$z_{3} = - \frac{51^{\frac{2}{3}}}{17} + \frac{3 \cdot 17^{\frac{2}{3}} \sqrt[6]{3} i}{17}$$
do backward replacement
$$z = x$$
$$x = z$$

$$x_{1} = \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$
$$x_{1} = \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$
This roots
$$x_{1} = \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$
=
$$- \frac{1}{10} + \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$
substitute to the expression
$$- 17 x x^{2} + 72 > 0$$
$$- 17 \left(- \frac{1}{10} + \frac{2 \cdot 51^{\frac{2}{3}}}{17}\right) \left(- \frac{1}{10} + \frac{2 \cdot 51^{\frac{2}{3}}}{17}\right)^{2} + 72 > 0$$

                     2                   
     /           2/3\                    
     |  1    2*51   |  /17       2/3\ > 0
72 + |- -- + -------| *|-- - 2*51   |    
     \  10      17  /  \10          /    

the solution of our inequality is:
$$x < \frac{2 \cdot 51^{\frac{2}{3}}}{17}$$

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