#"There is a method to solve a cubic equation in general by hand"# #"(and calculator) on paper. It is a method based on the substi-"#
#"tution of Vieta."#
#"Dividing by the first coefficient yields :"#
#x^3 - (17/4) x^2 - 1 = 0#
#"Substituting "x=y+p" in "x^3+ax^2+bx+c" yields :"#
#y^3 + (3p+a) y^2 + (3p^2+2ap+b) y + p^3+ap^2+bp+c = 0#
#"if we take "3p+a=0 => p=-a/3", the first coefficient becomes"# #"zero, and we get :"#
#y^3 - (289/48) y - (5777/864) = 0#
#"(with p = 17/12)"#
#"Substituting y=qz in "y^3 + b y + c = 0", yields :"#
#z^3 + b z / q^2 + c / q^3 = 0#
#"if we take "q = sqrt(|b|/3)", the coefficient of z becomes 3 or -3,"#
#"and we get :"#
#"(here q = 1.41666667)"#
#z^3 - 3 z - 2.35171993 = 0#
#"Substituting z = t + 1/t, yields :"#
#t^3 + 1/t^3 - 2.35171993 = 0#
#"Substituting "u = t^3", yields the quadratic equation :"#
#u^2 - 2.35171993 u + 1 = 0#
#"A root of this quadratic equation is u=1.79444436."#
#"Substituting the variables back, yields :"#
#t = root3(u) = 1.21518761.#
#=> z = 2.03810581.#
#=> y = 2.88731656.#
#=> x = 4.30398323.#
#"The other roots can be found by dividing and solving the"# #"remaining quadratic equation."#
#"The other roots are complex : "-0.02699161 pm 0.48126330 i."#
