#"There are no rational roots and no easy factorization."#
#"There is a method though to solve a cubic equation in general"# #"by hand (and calculator) on paper. This method that is used"#
#"here is based on the substitution of Vieta."#
#"Dividing by the first coefficient yields :"#
#x^3 - x^2 + (11/4) x - (3/2) = 0#
#"Substituting "x=y+p" in "x^3+ax^2+bx+c = 0" 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" or "p=-a/3", the first coefficient"#
#"becomes zero, and we get :"#
#y^3 + (29/12) y - (71/108) = 0#
#"(with "p = 1/3")"#
#"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 = 0.89752747")"#
#z^3 + 3 z - 0.90926683 = 0#
#"Substituting "z = t - 1/t", yields :"#
#t^3 - 1/t^3 - 0.90926683 = 0#
#"Substituting "u = t^3", yields the quadratic equation :"#
#u^2 - 0.90926683 u - 1 = 0#
#"A root of this quadratic equation is "u=1.55312854.#
#"Substituting the variables back, yields :"#
#t = root3(u) = 1.15807264.#
#=> z = 0.29456894.#
#=> y = 0.26438372.#
#=> x = 0.59771705.#
#"The other roots can be found by dividing and solving the"# #"remaining quadratic equation."#
#"The other roots are complex : "0.20114147 pm 1.57133406 i.#
