-9*x+x^3+3*x^2+5=0 equation

Given the equation:
$$\left(3 x^{2} + \left(x^{3} - 9 x\right)\right) + 5 = 0$$
transform
$$\left(- 9 x + \left(\left(3 x^{2} + \left(x^{3} - 1\right)\right) - 3\right)\right) + 9 = 0$$
or
$$\left(- 9 x + \left(\left(3 x^{2} + \left(x^{3} - 1^{3}\right)\right) - 3 \cdot 1^{2}\right)\right) + 9 = 0$$
$$- 9 \left(x - 1\right) + \left(3 \left(x^{2} - 1^{2}\right) + \left(x^{3} - 1^{3}\right)\right) = 0$$
$$- 9 \left(x - 1\right) + \left(\left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right) + 3 \left(x - 1\right) \left(x + 1\right)\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(3 \left(x + 1\right) + \left(\left(x^{2} + x\right) + 1^{2}\right)\right) - 9\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} + 4 x - 5\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} + 4 x - 5 = 0$$
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:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 4$$
$$c = -5$$
, then

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

(4)^2 - 4 * (1) * (-5) = 36

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

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

x3 = (-b - sqrt(D)) / (2*a)

or
$$x_{2} = 1$$
$$x_{3} = -5$$
The final answer for -9*x + x^3 + 3*x^2 + 5 = 0:
$$x_{1} = 1$$
$$x_{2} = 1$$
$$x_{3} = -5$$