In 1545 Cardano published Ars Magna, the first Latin treatise on algebra.

Cardano heard of Tartaglia's feat and petitioned him to share his findings so that they could be included, with all due credit, in the book Cardano was busy preparing. Wishing to see his discovery first published in one of his own forthcoming works, however, Tartaglia declined to divulge the secret.

At Cardano's subsequent entreaties, though, Tartaglia capitulated sometime in 1539. As he promised not to tell anyone he got the method concealed in a poem in Italic:

Italian	In free translation
Quando che'l cubo con le cose appresso	When the cube and the unknown together
Se agguaglia a qualche numero discreto:	is equal to some whole number,
	[ To solve x^3 + cx = d ]
Trovati dui altre differenti in esso.	Search for two other numbers which difference is equal to the number
Dapoi terrai, questo per consueto,	This shall You always do
Che'l loro produtto, sempre sia eguale	so that their product always is equal to
Al terzo cubo delle cose netto;	exactly one third of the number cubed
	[Find u, v so that u -v = d and uv =(c/3)^3]
El residuo poi suo generale,	In general it is so that the
Delle lor latti cubi, ben sottratti	difference between their cubes
Varra la tua cosa principale.	is what you are looking for
	[ x  =  ^3 u -  ^3 v ]
In el secondo, de cotesti atti;	In the other case,
Quando che'l cubo restasse lui solo,	when the cube is alone
	[ x^3 = cx + d ]
Tu osserverai quest' altri contratti,	You have to observe these other arrangement:
Del numer farai due, tal part'a volo	You split the number in two
Che'l una, in l'altra, si produca schietto,	so that one multiplied by the other gives
El terzo cubo delle cose in stolo;	exactly one third of the number, cubed.
	[ Find u, v so that u +v =d and uv = (c/3)^3 ]
Delle qual poi, per commun precetto,	In general it is so that
Torrai li lati cubi, insieme gionti,	the sum of their cubes
Et cotal somma, sara il tuo concetto;	is what You are looking for
	[ x  =  ^3 u +  ^3v ]
El terzo, poi di questi nostri conti,	The third of our calculations
Se solve nol secondo, se ben guardi	is solved in the same way if You take care,
Che per natura son quasi congionti.	as they are of the same kind.
	[ The third case, to solve x^3+ d = cx , is done in the same way as the second ]
Questi trovai, et non con passi tardi	This I have found, without heavy feet,
Nel mille cinquecent'e quattro e trenta;	This year, one thousand five hundred and thirty four.
Con fondamenti ben saldi e gagliardi	On safe ground
Nella Citta del mar intorno centa	In the town surrounded by sea.

But Cardano did not understand this poetic explanation and had to get more help from Tartaglia. The method Cardano then published in Ars Magna 1545, he first illustrated with the solution of the equation x³+6x=20.

Here, in modern notation, is the so called Cardan's solution of x³ + mx = n with m and n positive:

Cardano introduces two quantities t and u and lets t - u = n

and (tu) = (m/3)³

He then asserts that x =³(t) - ³(u)

By elimination , and solving the resulting quadratic equation, he obtains

t = ²[(n/2)²+ (m/3)³] + n/2, and

u = ²[(n/2)²+ (m/3)³] - n/2

and hence gets a value of x

By publishing the result in 1545, Cardano sparked one of the most spectacular priority controversies in the history of mathematics. Of importance is the fact that Cardano's publication of solutions of the cubic equations brought these variations into competition within the mathematical environment.

Cardano noticed something strange when he applied his formula to certain cubics. When solving x³ = 15x + 4 he obtained an expression involving (-121). Cardan knew that you couldn't take the square root of a negative number yet he also knew that x = 4 was a solution to the equation. He wrote to Tartaglia on 4 August 1539 in an attempt to clear up the difficulty. Tartaglia certainly did not understand the problem. In Ars Magna Cardan gives a calculation with 'complex numbers' to solve a similar problem but he really did not understand his own calculation which he says is as subtle as it is useless.

Raphael Bombelli 1526-1573 was the one who finally cleared out these difficulties with imaginary numbers. In Algebra 1569 , Bombelli solves equations, using the method of del Ferro/Tartaglia, introduces +i and -i and describes what happens when these are multiplied in various combinations.

x³ = 15x +4 has the real root x=4 which easily can be seen by testing x = 4.

 But solving the equation using the method of del Ferro/Tartaglia leads to:

(u+v) ³ =15(u+v) +4 gives u³ +v³ = 4 and u³*v³ =125

which gives vp +q = 4 and p*q =125 which has
the solution p =2+ (-121) and q =2- (-121)

Hence   x = ³p-³q = ³(2 + (-121)) +  ³(2 - (-121))  =
= ³(2 + 11i) +³(2-11i) and we know that this expression is   = 4. To show this requires quite a lot of work, and some cleverness.

It is reasonable to start with assuming that    2+11i  can be expressed as a cube.
And it can easily be shown that this is true as 2 +11i = (2 + i) ³ and also 
that 2 - 11i = (2 - i) ³ and hence: (2 - i) ³ = 8 -3*4i-3*2*1+i = 2-11i

And (2 + i) ³ = 8+3*4i-3*2*1- i = 2+11i

And we finally get = ((2+i) ³ ) ^1/3 + ((2-i) ³ ) ^1/3 = 2+2=4

Another way of reaching an understanding of this can be shown by examining the  equation   x³ + 6x = 20. Obviously there is an answer  x = 2, but using the formula of Tartaglia leads to the answer x =  ³((108) +10) - ³( (108) -10) which then obviously must be equal to 2.

A way of solving this is by regarding (a+b)³ = 10 + 108   and   (a - b)³ = 10 - 108
so   a³ +3ab²= 10 ,  3a²b+ b³ = 108 and since (a+b) + (a-b) is to be 2, we should have  a=1, so the  equations become:  1 + 3b² = 10    and    3b + b³ = 108 = 63
which do have the common solution b = 3 so all that is required is to check that 
(1 + 3)³ = 10 +108, whence of course  (1  - 3)³ = 10 - 108

Raphael Bombelli managed to do this and this made it possible for him to work out a theory for imaginary numbers. In the years after Cardan's Ars Magna many mathematicians contributed to the solution of cubic equations. Viète, Harriot, Tschirnhaus, Euler, Bezout and Descartes all devised methods. Tschirnhaus' methods were extended by the Swedish mathematician E S Bring near the end of the 1800 C.

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