MATHS!!!!

Numbers that are multiples of the square root of minus one are called "complex numbers".

Complex number - Wikipedia

1 2 3 5 8 13 21 34 55 89 144 233 377 610 987

fibonacci sequence lol

Unless it's -8.

i hate maths

In algebra, a cubic equation in one variable is an equation of the form in which a is nonzero. The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. If all of the coefficients a, b, c, and d of the cubic equation are real numbers, then it has at least one real root (this is true for all odd-degree polynomial functions). All of the roots of the cubic equation can be found by the following means: algebraically, that is, they can be expressed by a cubic formula involving the four coefficients, the four basic arithmetic operations and nth roots (radicals). (This is also true of quadratic (second-degree) and quartic (fourth-degree) equations, but not of higher-degree equations, by the Abel–Ruffini theorem.) trigonometrically numerical approximations of the roots can be found using root-finding algorithms such as Newton's method. The coefficients do not need to be real numbers. Much of what is covered below is valid for coefficients in any field with characteristic other than 2 and 3. The solutions of the cubic equation do not necessarily belong to the same field as the coefficients. For example, some cubic equations with rational coefficients have roots that are irrational (and even non-real) complex numbers. Cubic equations were known to the ancient Babylonians, Greeks, Chinese, Indians, and Egyptians.[1][2][3] Babylonian (20th to 16th centuries BC) cuneiform tablets have been found with tables for calculating cubes and cube roots.[4][5] The Babylonians could have used the tables to solve cubic equations, but no evidence exists to confirm that they did.[6] The problem of doubling the cube involves the simplest and oldest studied cubic equation, and one for which the ancient Egyptians did not believe a solution existed.[7] In the 5th century BC, Hippocrates reduced this problem to that of finding two mean proportionals between one line and another of twice its length, but could not solve this with a compass and straightedge construction,[8] a task which is now known to be impossible. Methods for solving cubic equations appear in The Nine Chapters on the Mathematical Art, a Chinese mathematical text compiled around the 2nd century BC and commented on by Liu Hui in the 3rd century.[2] In the 3rd century AD, the Greek mathematician Diophantus found integer or rational solutions for some bivariate cubic equations (Diophantine equations).[3][9] Hippocrates, Menaechmus and Archimedes are believed to have come close to solving the problem of doubling the cube using intersecting conic sections,[8] though historians such as Reviel Netz dispute whether the Greeks were thinking about cubic equations or just problems that can lead to cubic equations. Some others like T. L. Heath, who translated all of Archimedes' works, disagree, putting forward evidence that Archimedes really solved cubic equations using intersections of two conics, but also discussed the conditions where the roots are 0, 1 or 2.[10]  In the 7th century, the Tang dynasty astronomer mathematician Wang Xiaotong in his mathematical treatise titled Jigu Suanjing systematically established and solved numerically 25 cubic equations of the form x3 + px2 + qx = N, 23 of them with p, q ≠ 0, and two of them with q = 0.[11] In the 11th century, the Persian poet-mathematician, Omar Khayyam (1048–1131), made significant progress in the theory of cubic equations. In an early paper, he discovered that a cubic equation can have more than one solution and stated that it cannot be solved using compass and straightedge constructions. He also found a geometric solution.[12][13] In his later work, the Treatise on Demonstration of Problems of Algebra, he wrote a complete classification of cubic equations with general geometric solutions found by means of intersecting conic sections.[14][15] In the 12th century, the Indian mathematician Bhaskara II attempted the solution of cubic equations without general success. However, he gave one example of a cubic equation: x3 + 12x = 6x2 + 35.[16] In the 12th century, another Persian mathematician, Sharaf al-Dīn al-Tūsī (1135–1213), wrote the Al-Muʿādalāt (Treatise on Equations), which dealt with eight types of cubic equations with positive solutions and five types of cubic equations which may not have positive solutions. He used what would later be known as the "Ruffini-Horner method" to numerically approximate the root of a cubic equation. He also used the concepts of maxima and minima of curves in order to solve cubic equations which may not have positive solutions.[17] He understood the importance of the discriminant of the cubic equation to find algebraic solutions to certain types of cubic equations.[18] In his book Flos, Leonardo de Pisa, also known as Fibonacci (1170–1250), was able to closely approximate the positive solution to the cubic equation x3 + 2x2 + 10x = 20. Writing in Babylonian numerals he gave the result as 1,22,7,42,33,4,40 (equivalent to 1 + 22/60 + 7/602 + 42/603 + 33/604 + 4/605 + 40/606), which has a relative error of about 10−9.[19] In the early 16th century, the Italian mathematician Scipione del Ferro (1465–1526) found a method for solving a class of cubic equations, namely those of the form x3 + mx = n. In fact, all cubic equations can be reduced to this form if one allows m and n to be negative, but negative numbers were not known to him at that time. Del Ferro kept his achievement secret until just before his death, when he told his student Antonio Fior about it. Niccolò Fontana Tartaglia In 1530, Niccolò Tartaglia (1500–1557) received two problems in cubic equations from Zuanne da Coi and announced that he could solve them. He was soon challenged by Fior, which led to a famous contest between the two. Each contestant had to put up a certain amount of money and to propose a number of problems for his rival to solve. Whoever solved more problems within 30 days would get all the money. Tartaglia received questions in the form x3 + mx = n, for which he had worked out a general method. Fior received questions in the form x3 + mx2 = n, which proved to be too difficult for him to solve, and Tartaglia won the contest. Later, Tartaglia was persuaded by Gerolamo Cardano (1501–1576) to reveal his secret for solving cubic equations. In 1539, Tartaglia did so only on the condition that Cardano would never reveal it and that if he did write a book about cubics, he would give Tartaglia time to publish. Some years later, Cardano learned about del Ferro's prior work and published del Ferro's method in his book Ars Magna in 1545, meaning Cardano gave Tartaglia six years to publish his results (with credit given to Tartaglia for an independent solution). Cardano's promise to Tartaglia said that he would not publish Tartaglia's work, and Cardano felt he was publishing del Ferro's, so as to get around the promise. Nevertheless, this led to a challenge to Cardano from Tartaglia, which Cardano denied. The challenge was eventually accepted by Cardano's student Lodovico Ferrari (1522–1565). Ferrari did better than Tartaglia in the competition, and Tartaglia lost both his prestige and his income.[20] Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with these complex numbers in Ars Magna, but he did not really understand it. Rafael Bombelli studied this issue in detail[21] and is therefore often considered as the discoverer of complex numbers. François Viète (1540–1603) independently derived the trigonometric solution for the cubic with three real roots, and René Descartes (1596–1650) extended the work of Viète.[22] If the coefficients of a cubic equation are rational numbers, one can obtain an equivalent equation with integer coefficients, by multiplying all coefficients by a common multiple of their denominators. Such an equation with integer coefficients, is said to be reducible if the polynomial of the left-hand side is the product of polynomials of lower degrees. By Gauss's lemma, if the equation is reducible, one can suppose that the factors have integer coefficients. Finding the roots of a reducible cubic equation is easier than solving the general case. In fact, if the equation is reducible, one of the factors must have the degree one, and have thus the form with q and p being coprime integers. The rational root test allows finding q and p by examining a finite number of cases (because q must be a divisor of a, and p must be a divisor of d). Thus, one root is  and the other roots are the roots of the other factor, which can be found by polynomial long division. This other factor is (The coefficients seem not to be integers, but must be integers if p / q is a root.) Then, the other roots are the roots of this quadratic polynomial and can be found by using the quadratic formula. Cubics of the form are said to be depressed. They are much simpler than general cubics, but are fundamental, because the study of any cubic may be reduced by a simple change of variable to that of a depressed cubic. Let be a cubic equation. The change of variable gives a cubic (in t) that has no term in t2. After dividing by a one gets the depressed cubic equation with The roots  of the original equation are related to the roots  of the depressed equation by the relations for .

It's obviously just a copy and paste. You can tell by the footnotes "[1][2][3]".

Probably from the wiki.