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

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.

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

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

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


 * $u = \begin{cases}

& S \\ \left ({-\dfrac 1 2 + \dfrac {i \sqrt 3} 2}\right) & S \\ \left ({-\dfrac 1 2 - \dfrac {i \sqrt 3} 2}\right) & S \\ \end{cases}$


 * $v = \begin{cases}

& T \\ \left ({-\dfrac 1 2 + \dfrac {i \sqrt 3} 2}\right) & T \\ \left ({-\dfrac 1 2 - \dfrac {i \sqrt 3} 2}\right) & 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 \\ \left ({-\dfrac 1 2 + \dfrac {i \sqrt 3} 2}\right) S + \left ({-\dfrac 1 2 - \dfrac {i \sqrt 3} 2}\right) T = & -\dfrac {S + T} 2 + \dfrac {i \sqrt 3} 2 \left({S - T}\right)\\ \left ({-\dfrac 1 2 - \dfrac {i \sqrt 3} 2}\right) S + \left ({-\dfrac 1 2 + \dfrac {i \sqrt 3} 2}\right) T = & -\dfrac {S + T} 2 - \dfrac {i \sqrt 3} 2 \left({S - T}\right)\\ \end{cases}$

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


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

This method (in an incomplete form) was first published by Gerolamo Cardano in 1545. He learned the technique from Niccolò Fontana Tartaglia, who had sworn him to secrecy. However, as Cardano learned in 1543, the technique had in fact first been discovered by Scipione del Ferro, so he no longer felt bound by his oath to Tartaglia. The latter did not see the matter in the same light, and entered into a feud with Cardano that lasted a decade.

The method detailed here was not actually analyzed in depth until the work of Rafael Bombelli, who was the first one to solve the problem of what to do about the "imaginary numbers" that inevitably arose when using this formula.

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