Cardano's Formula

Theorem
Let $P$ be the cubic equation:
 * $a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$

Then $P$ has solutions:
 * $x_1 = S + T - \dfrac b {3 a}$
 * $x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$
 * $x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$

where:
 * $S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
 * $T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$

where:
 * $Q = \dfrac {3 a c - b^2} {9 a^2}$
 * $R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$

The expression $D = Q^3 + R^2$ is called the discriminant of the equation.

Real Coefficients
Let $a, b, c, d \in \R$.

Then:

Proof
First the cubic is depressed, by using the Tschirnhaus Transformation:
 * $x \to x + \dfrac b {3 a}$:

Now let:
 * $y = x + \dfrac b {3 a}, Q = \dfrac {3 a c - b^2} {9 a^2}, R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$

Thus we have obtained the depressed cubic $y^3 + 3 Q y - 2 R = 0$.

Now let $y = u + v$ where $u v = -Q$.

Then:

Thus the resolvent equation is obtained.

This resolvent is seen to be a quadratic in $u^3$.

From Solution to Quadratic Equation:


 * $u^3 = \dfrac {2 R \pm \sqrt {4 Q^3 + 4 R^2}} 2 = R \pm \sqrt {Q^3 + R^2}$

We have from above $u v = -Q$ and hence $v^3 = -\dfrac {Q^3} {u^3}$.

Let us try taking the positive root: $u^3 = R + \sqrt {Q^3 + R^2}$.

Then:

The same sort of thing happens if you start with $u^3 = R - \sqrt {Q^3 + R^2}$: we get $v^3 = R + \sqrt {Q^3 + R^2}$.

Thus we see that taking either square root we arrive at the same solution.


 * $u^3 = R + \sqrt {Q^3 + R^2}$
 * $v^3 = R - \sqrt {Q^3 + R^2}$
 * $v^3 = R - \sqrt {Q^3 + R^2}$

Let:
 * $S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
 * $T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$

From Roots of Complex Number, we have the three cube roots of $u^3$ and $v^3$:


 * $u = \begin{cases}

& S \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & S \\ \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & S \\ \end{cases}$


 * $v = \begin{cases}

& T \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} & T \\ \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} & T \\ \end{cases}$

Because of our constraint $u v = -Q$, there are only three combinations of these which are possible such that $y = u + v$:


 * $ y = \begin{cases}

& S + T \\ \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \paren {S - T} \\ \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2} S + \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2} T = & -\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \paren {S - T} \\ \end{cases}$

As $y = x + \dfrac b {3a}$, it follows that the three roots are therefore:


 * $(1): \quad x_1 = S + T - \dfrac b {3 a}$
 * $(2): \quad x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$
 * $(3): \quad x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$

Also known as
Cardan's Formula, from the English form of 's name, Jerome Cardan.

Some use the term Cardano's method.

Some sources refer to it as the Tartaglia formula, acknowledging the work of in its development.

Also see

 * Tartaglia's Poem