Cubic Equation Solver

Solve ax³ + bx² + cx + d = 0. Enter four coefficients to get the discriminant and all roots — three real or one real plus two complex conjugates.

Inputs

ax^3 + bx^2 + cx + d = 0

Results

xy
\begin{aligned} p &= \dfrac{3ac - b^2}{3a^2} \\ &= \dfrac{3\left(1\right)\left(11\right) - \left(-6\right)^2}{3\left(1\right)^2} \\ &= ? \end{aligned}
\begin{aligned} q &= \dfrac{2b^3 - 9abc + 27a^2 d}{27a^3} \\ &= \dfrac{2\left(-6\right)^3 - 9\left(1\right)\left(-6\right)\left(11\right) + 27\left(1\right)^2\left(-6\right)}{27\left(1\right)^3} \\ &= ? \end{aligned}
\begin{aligned} \Delta &= -4p^3 - 27q^2 \\ &= -4\left(?\right)^3 - 27\left(?\right)^2 \\ &= ? \end{aligned}
\begin{aligned} x_1 &= 2r\cos\!\left(\theta\right) - h \\ &= 2\left(?\right)\cos\!\left(?\right) - \left(?\right) \\ &= ? \end{aligned}
\begin{aligned} x_2 &= 2r\cos\!\left(\theta - \dfrac{2\pi}{3}\right) - h = ? \end{aligned}
\begin{aligned} x_3 &= 2r\cos\!\left(\theta - \dfrac{4\pi}{3}\right) - h = ? \end{aligned}

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