the two factors. So the possible values of are: or Using another from among the three roots of the cubic simply causes these same four values of to appear in a different order. The solutions of the cubic are: using any one of the three possible cube roots. A wise strategy is to choose the sign of the square-root that makes the absolute value of as large as possible. Ferrari's solution Otherwise, the depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity to equation (), yielding The effect has been to fold up the u4 term into a perfect square: (u2 + α)2. The second term, αu2 did not disappear, but its sign has changed and it has been moved to the right side. The next step is to insert a variable y into the perfect square on the left side of equation (), and a corresponding 2y into the coefficient of u2 in the right side. To accomplish these insertions, the following valid formulas will be added to equation (), and These two formulas, added together, produce which added to equation () produces This is equivalent to The objective now is to choose a value for y such that the right side of equation () becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function: The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero: Therefore to make the right side of equation () into a perfect square, the following equation must be solved: Multiply the binomial with the polynomial, Divide both sides by −4, and move the −β2/4 to the right, Divide both sides by 2, This is a cubic equation in y. Solve for y using any method for solving such equations (e.g. conversion to a reduced cubic and application of Cardano's formula). Any of the three possible roots will do. Folding the second perfect square With the value for y so selected, it is now known that the right side of equation () is a perfect square of