Cubic Equation Formula (Cardano) with Examples

Key Points

• The cubic equation formula applies to the depressed cubic $t^3 + pt + q = 0$ and gives $t = \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$.

• Any cubic must first be depressed by substituting $x = t - \frac{b}{3a}$, which removes the squared term.

• MacTutor records that Cardano published methods for solving the cubic and quartic in Ars Magna in 1545, after obtaining the cubic method from Tartaglia in 1539.

• The discriminant $\Delta = -4p^3 - 27q^2$ decides the root pattern: positive gives three real roots, zero a repeated root, negative one real and two complex.

• When three real roots exist the formula requires cube roots of complex numbers, a situation known as the casus irreducibilis.

What Is the Cubic Equation Formula?

The cubic equation formula solves a cubic that has had its squared term removed. For $t^3 + pt + q = 0$:

$$t = \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} \ + \ \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$

This article covers that solution method. For the structure and factoring of cubic expressions, see cubic polynomials and factoring cubic polynomials, which handle those separately.

Why the Cubic Must Be Depressed First

The formula has no place for a squared term, so a general cubic $ax^3+bx^2+cx+d=0$ cannot be fed into it directly. Substituting $x = t - \frac{b}{3a}$ shifts the variable and makes the $t^2$ coefficient vanish.

The shift works because the sum of the three roots equals $-\frac{b}{a}$, so moving the origin to their average centres them and kills the squared term. What remains is $t^3+pt+q=0$ in the new variable.

After solving for $t$, the shift must be undone with $x = t - \frac{b}{3a}$. Forgetting that last step returns a root of the depressed cubic rather than of the original equation.

Variable Key

Example 1

Solve $t^3 + 3t - 4 = 0$ using the formula.

Here $p = 3$ and $q = -4$, so $R = \frac{16}{4} + \frac{27}{27} = 4 + 1 = 5$:

$$t = \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} \approx 1.618 - 0.618 = 1$$

Final answer: $t = 1$, which checks since $1 + 3 - 4 = 0$.

Example 2

Solve $x^3 - 6x^2 + 11x - 6 = 0$ by applying the formula to it directly.

The formula looks ready to use with $p = 11$ and $q = -6$ taken straight from the equation. Doing that gives $R = 9 + \frac{1331}{27} \approx 58.3$, and the resulting value does not satisfy the original equation.

The reason is structural: the formula solves $t^3+pt+q=0$, which has no squared term, and this equation has $-6x^2$. The coefficients $p$ and $q$ do not exist until the cubic is depressed.

Substituting $x = t + 2$ removes the squared term:

$$t^3 - t = 0 \implies t(t-1)(t+1) = 0 \implies t = 0, 1, -1$$ $$x = t + 2 \implies x = 2, 3, 1$$

Final answer: $x = 1, 2, 3$. Here the rational root theorem would have found these faster, which is usually true when the roots are small integers.

Example 3

Solve $t^3 - 15t - 4 = 0$ and comment on the intermediate values.

With $p = -15$ and $q = -4$, $R = 4 - 125 = -121$, so $\sqrt{R} = 11i$:

$$t = \sqrt[3]{2+11i} + \sqrt[3]{2-11i} = (2+i) + (2-i) = 4$$

Final answer: $t = 4$, since $64 - 60 - 4 = 0$.

The root is a plain integer, yet the formula reached it through complex numbers. This is the casus irreducibilis, and it is what pushed sixteenth-century algebraists towards accepting complex quantities.

Example 4

Use the discriminant to describe the roots of $t^3 - 7t + 6 = 0$.

$$\Delta = -4(-7)^3 - 27(6)^2 = 1372 - 972 = 400$$

Final answer: $\Delta > 0$, so the equation has three distinct real roots. They are $1$, $2$ and $-3$.

Example 5

Use the discriminant to describe the roots of $t^3 + t + 1 = 0$.

$$\Delta = -4(1)^3 - 27(1)^2 = -4 - 27 = -31$$

Final answer: $\Delta < 0$, so there is one real root and two complex conjugate roots.

A negative discriminant is the comfortable case for the formula, because $R$ is then positive and no complex cube roots are needed.

When to Use the Formula and When Not To

The formula is a guarantee rather than a first resort. It always works, but for school-level cubics the rational root theorem reaches the answer in a fraction of the effort.

Cubic Equation, Where Things Go Sideways

Mistake 1: Skipping the depression step

Where it slips in: The formula names $p$ and $q$, and a general cubic has coefficients that look like they could serve.

Don't do this: Read $p$ and $q$ off $ax^3+bx^2+cx+d$ while $b \neq 0$.

The correct way: Substitute $x = t - \frac{b}{3a}$ first, then read $p$ and $q$ from the depressed form.

Students who check whether the squared term is absent before writing anything down never make this error. One glance at the $x^2$ coefficient decides whether the formula is even applicable yet.

Mistake 2: Forgetting to undo the substitution

Where it slips in: The formula returns a value and it feels like the answer.

Don't do this: Report the value of $t$ as a root of the original equation.

The correct way: Convert back with $x = t - \frac{b}{3a}$. In Example 2 the depressed roots were $0, 1, -1$ while the real answers were $2, 3, 1$.

Mistake 3: Reaching for the formula before trying rational roots

Where it slips in: The question says cubic, so the cubic formula seems to be what is wanted.

Don't do this: Run Cardano's method on $x^3-6x^2+11x-6=0$.

The correct way: Test small divisors of the constant term first. That equation factors as $(x-1)(x-2)(x-3)$ in a couple of lines, and the formula would take many more.

Bottom Line

• The cubic equation formula solves $t^3+pt+q=0$ as the sum of two cube roots.

• A general cubic must be depressed by $x = t - \frac{b}{3a}$ before the formula applies, and the shift must then be undone.

• The discriminant $-4p^3-27q^2$ predicts the root pattern before any roots are computed.

• Three real roots force the formula through complex intermediate values, the casus irreducibilis.

• To work through polynomial equations with a teacher, explore an algebra tutor or math tuition.

Where to Go From Here

Solve $t^3 - 6t - 9 = 0$ with the formula. Then take $x^3 - 3x^2 - 4x + 12 = 0$, find a rational root by inspection, and compare how much work each route costs.

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Read More

• Linear, Quadratic and Cubic Polynomials

• Degree of a Polynomial

• Discriminant

• Rational Root Theorem

• Dividing Polynomials