the form (This is correct for both signs of square root, as long as the same sign is taken for both square roots. A ± is redundant, as it would be absorbed by another ± a few equations further down this page.) so that it can be folded: Note: If β ≠ 0 then α + 2y ≠ 0. If β = 0 then this would be a biquadratic equation, which we solved earlier. Therefore equation () becomes Equation () has a pair of folded perfect squares, one on each side of the equation. The two perfect squares balance each other. If two squares are equal, then the sides of the two squares are also equal, as shown by: Collecting like powers of u produces Note: The subscript s of and is to note that they are dependent. Equation () is a quadratic equation for u. Its solution is Simplifying, one gets This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are Remember: The two come from the same place in equation (), and should both have the same sign, while the sign of is independent. Summary of Ferrari's method Given the quartic equation its solution can be found by means of the following calculations: If then Otherwise, continue with (either sign of the square root will do) (there are 3 complex roots, any one of them will do) The two ±s must have the same sign, the ±t is independent. To get all roots, compute x for ±s,±t = +,+ and for +,−; and for −,+ and for −,−. This formula handles repeated roots without problem. Ferrari was the first to discover one of these labyrinthine solutions. The equation which he solved was which was already in depressed form. It has a pair of solutions which can be found with the set of formulas shown above. Ferrari's solution in the special case of real coefficients If the coefficients of the quartic equation are real then the nested depressed cubic equation () also has real coefficients, thus it has at least one real root. Furthermore the cubic function where P and Q are given by () has the properties that and where α and β are given by (). This means that () has a real root greater than , and therefore that () has a real root greater than . Using this