x^3-9x^2+26x-24=0 equation

Given the equation:
$$x^{3} - 9 x^{2} + 26 x - 24 = 0$$
transform
$$x^{3} - 9 x^{2} + 26 x - 24 = 0$$
or
$$x^{3} + 26 x - 60 = 0$$
$$x^{3} - 9 x^{2} + 26 x - 24 = 0$$
$$\left(- 9 x + 18\right) \left(x + 2\right) + \left(x - 2\right) \left(x^{2} + 2 x + 4\right) + 26 x - 52 = 0$$
Take common factor $x - 2$ from the equation
we get:
$$\left(x - 2\right) \left(x^{2} - 7 x + 12\right) = 0$$
or
$$\left(x - 2\right) \left(x^{2} - 7 x + 12\right) = 0$$
then:
$$x_{1} = 2$$
and also
we get the equation
$$x^{2} - 7 x + 12 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -7$$
$$c = 12$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 12 + \left(-7\right)^{2} = 1$$
Because D > 0, then the equation has two roots.
$$x_2 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_3 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{2} = 4$$
Simplify
$$x_{3} = 3$$
Simplify
The final answer for (x^3 - 9*x^2 + 26*x - 1*24) + 0 = 0:
$$x_{1} = 2$$
$$x_{2} = 4$$
$$x_{3} = 3$$