In the book Galois theory by J. P. Escofier, author wrote
In his treatise on Algebra (from around $1074$), Omar Khayyam studied cubic equations in detail. He only considered equations with positive coefficients, and distinguished $25$ different cases. For example, the equations with three terms not having $0$ as a root are one of the following six types: $$ x^3=ax^2+b, \,\,\,\,\,\, x^3+b=ax^2, \,\,\,\,\,\, x^3+ax^2=b, $$ $$ x^3=ax+b, \,\,\,\,\,\, x^3+b=ax, \,\,\,\,\,\, x^3+ax=b. $$
After around $500$ years:
Scipio del Ferro discovered solutions of equations $$(*) \,\,\,\,\,\,\,\,\,\,\,\,\,\, x^3+px=q, \,\,\,\, x^3=px+q, \,\,\,\, x^3+q=px$$ where $p,q>0$. .... In $1535$, Fiore (student of Ferro) challenged Tartaglia to solve 30 problems, all based on equations of first type above. ... Tartaglia had already attempted to solve equations of this type some years earlier, so he succeeded (to find answers to problems of Fiore, i.e. to solve cubic equations of type $x^3+px=q$ with $p,q>0$.) ............In $1539$, Cardan invited Tartaglia to find out his (Tartaglia's) secret (of solution of cubic). He flattered him so well that he succeeded - Tartaglia showed him (secret in) his poem, but swore not to reveal it. Shortly after, Cardan succeeded in extending Tartaglia's method to equations of second and third type in $(*)$ ......... In $1545$, Cardan published all of these solutions. But Tartaglia was furious, denounced him for lying .......
Question: My question is historical. If we type Tartaglia-Cardano in Google, we can see a story, as in quoted paragraph above. But, if the solutions of cubic equations were known to Omar Khayyam in $1074$, why Tartaglia and Cardano were curious/worried about solution of cubic equations and its publication?
Khayyam's solutions are geometric constructions (remember this new field of algebra needed to rest its foundation at the time on geometry in the style of Euclid's Elements book II, e.g. Proposition II-6 for a particular class of quadratic equations) that is roughly speaking, intersecting a hyperbola (or parabola, IIRC he didn't need ellipses) with a circle when the roots are not constructible (with straightedge and compass). That is not the more numerical formula (that is basically our modern day cubic formula) that Tartaglia, Fior, Cardano, Ferrari, etc. developed in the 16th century for the cubic and very soon after, quartic.
That is why you get those challenges in the open in this time. They actually had to solve for numerical values for given coefficients (so most people in the crowd could actually understand what was going on, who's right, and exciting to watch), rather than to say: "if we have this hyperbola and intersect with this circle, then this segment will represent the solution".