Solve any cubic equation of the form ax³ + bx² + cx + d = 0 by entering your coefficients a, b, c, and d. The calculator finds all real and complex roots using Cardano's formula, displaying up to three solutions for x — including repeated roots and complex pairs.
Root x₁
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Root x₂
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Root x₃
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Discriminant (Δ)
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Root Type
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A cubic equation is a polynomial equation of degree 3, written in standard form as ax³ + bx² + cx + d = 0, where a ≠ 0. The highest power of the variable x is 3. Cubic equations always have at least one real root, and can have up to three real roots.
The calculator uses Cardano's formula — a closed-form algebraic method for finding the roots of a cubic polynomial. It first converts the equation to a depressed cubic (no x² term) using a substitution, then applies the formula to extract all three roots, including any complex ones.
A cubic equation always has exactly three roots (counting multiplicity) in the complex number system. In the real number system, there is either one real root and two complex conjugate roots, or three real roots (which may include repeated values).
If a term is absent from your equation, enter 0 for its coefficient. For example, if your equation is x³ + 5x - 6 = 0, enter a = 1, b = 0, c = 5, d = -6. The coefficient 'a' must never be 0, otherwise the equation is not cubic.
The discriminant Δ = 18abcd − 4b³d + b²c² − 4ac³ − 27a²d² tells you about the nature of the roots. If Δ > 0, there are three distinct real roots. If Δ = 0, there is a repeated root. If Δ < 0, there is one real root and two complex conjugate roots.
Complex roots involve the imaginary unit i (√−1) and always appear as conjugate pairs, e.g. 2 + 3i and 2 − 3i. When a cubic has a negative discriminant, it yields one real root and two complex conjugate roots. This calculator displays the real part and imaginary part for complex solutions.
Yes, the leading coefficient 'a' can be any non-zero real number — positive or negative. A negative 'a' simply means the cubic curve opens downward on the left and upward on the right, but the solving method remains identical.
Substitute each root value back into ax³ + bx² + cx + d. If the result is 0 (or very close to 0 for rounded decimals), the root is correct. You can also verify by factoring: if x₁, x₂, x₃ are the roots, then a(x − x₁)(x − x₂)(x − x₃) should expand back to your original polynomial.