We will solve (for X) the polynomial of the form:
aX³ + bX² + cX + d = 0.
First input the rational coefficients for a, b, c, d.
(Click 'Next' when you're finished)

a : b : c : d :

Next

By dividing the coefficients by the value of 'a' we get,

X³ + X² + X + = 0

Next

We now transform the cubic by making the substitution Y=X+b/3 to put the polynomial in the form of Y³+pY+q = 0. Giving us

Y³ + Y + = 0

Next

We can now calculate the discriminant D = q²/4 + p³/27 or

D =

Next

We can now calculate the real root of Y³+pY+q = 0 by
thus

Y₁ =
By undoing the substitution done earlier we get X = Y₁ - B/3
X₁ =
And imaginary roots are calculated by

where w = (-1 + -3^1/2)/2
then we have the imaginary roots
Y₂ = + i
and
Y₃ = + i
and by undoing the original substitution we have
X₂ = + i
and
X₃ = + i