While the [quadratic formula] for solving [quadratic] equations is familiar to everyone who has taken high school math, it seems that the formula for solving [cubic] equations is not as familiar, but it is occasionally very handy. This formula was first discovered by [Girolamo Cardano], [Scipione del Ferro] and [Niccolo Fontana Tartaglia], and first published in Cardano's [Ars Magna] in 1545, and is considered a major result in [Renaissance] mathematics.

The general cubic equation can be expressed as follows:

x³ + bx² + cx + d = 0

We may then calculate the following two values (somewhat analogous to the [discriminant] of the quadratic formula):

q = (b² - 3c)/9

r = (2b³ 9bc + 27d)/54

If the [coefficient]s of the cubic are all real, then the sum p = q³ + r² determines what sort of roots the cubic polynomial might possess. If p is greater than 0, then it possesses one real root and two [complex conjugate] roots. If p is equal to zero, all roots are real, and we have roots of at least [multiplicity] 2. If p is less than zero, then all three roots are real.

In the latter case, we can proceed by computing:

θ = cos^-1(r/sqrt(q³))

The three real roots may then be readily computed as:

x₁ = -2 sqrt(q) cos(θ/3) - b/3

x₂ = -2 sqrt(q) cos((θ + 2π)/3) - b/3

x₃ = -2 sqrt(q) cos((θ - 2π/3) - b/3

If the coefficients are complex, or if either of the other two cases applies, we then compute these two values:

s₁ = (r + sqrt(q³ + r²))^1/3

s₂ = (r - sqrt(q³ + r²))^1/3

The roots of the cubic are then:

x₁ = s₁ + s₂ - b/3

x₂ = -(s₁ + s₂)/2 - b/3 + (i sqrt(3)/2)(s₁ - s₂)

x₃ = -(s₁ + s₂)/2 - b/3 - (i sqrt(3)/2)(s₁ - s₂)

In practice, it's usually easier to just use one of these formulas to compute the root (the first one usually), and then remove that root using [synthetic division] and solve the [resultant] quadratic equation.

If one is doing these kinds of calculations by hand, a simple check would be to use the [symmetric function]s generated by the cubic, namely:

x₁ + x₂ + x₃ = -b

x₁x₂ + x₁x₃ + x₂x₃ = c

x₁x₂x₃ = -d

and see if these formulas on the roots match with the coefficients of the original cubic.

In the [computer age] however, this cubic formula is only seldom used to numerically compute the roots of cubic polynomials, as iterative [numerical method]s like [Laguerre's method] or the [Jenkins-Traub method] can generally do the job much more efficiently for equations of this degree and higher.

Sources:

William H. Press, et. al. [Numerical Recipes] in FORTRAN

Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions

cubic formula (idea)