How do you solve a cubic equation using Tartaglias method?
You can find the actual formula for the solution to the cubic equation in many textbooks and online references. The link below is one. I’ll leave the actual calculations to you. However, it helps to know what you’re trying to accomplish instead of just plugging numbers into a formula. The first step is to substitute x = u+c for an appropriately selected constant c. The goal is to choose a constant that makes the coefficient of the u^2 term = 0. That leads to a (somewhat) simpler cubic equation u^3 + au + b = 0 It’s still not obvious how to make progress, but Tartaglia and one or two of his contemporaries noticed that making a second substitution, u = d(v + 1/v) with a properly selected constant d would cause enough additional simplification to generate an equation of the form v^6 + kv^3 + p = 0 This equation, conveniently enough, happens to be a quadratic equation in v^3. Therefore it can be solved for v^3 using the quadratic formula. Take cube roots to get the solutions for v and work
