R^3-3r^2-18r=0

Solution for R^3-3r^2-18r=0 equation:

^3-3R^2-18R=0

We add all the numbers together, and all the variables

-3R^2-18R=0

a = -3; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·(-3)·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$R_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*-3}=\frac{0}{-6}
=0
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*-3}=\frac{36}{-6}
=-6
$