I Solving complex formulas and higher order polynomials

It is not so easy to solve these cubics in practice. In good cases the solutions are integers or fractions, in which case they can be found by trial and error from the rational roots theorem taught in precalculus. This does not work in your example, unless i made a mistake, or at least they do not seem to be integers. In the general case the roots can be expressed using square roots and cube roots, by the method of scipio del ferro, often called incorrectly that of cardano. There are several steps involved, the first being a change of variable which makes the x^2 term equal to zero. Then when the equation has been r3ndered into the form X^3 = pX + q, one can show that a solution has the form X = u+v, provided u^3+v^3=q and 3uv = p. Since these equtions imply we know the product and the sum of the quantities u^3 and v^3, we use the fact that given the sum and product of two quantities, those quantities are found from solving a quadratic equation. I.e. we can write down the quadratic equation t^2 - qt + p/3 = 0, and solve it for u^3 and v^3. Then we can take cube roots to get u, and set v = p/3u.

Thjis is explained in section IV, (see page 253, and 264), of the excellent algebra book of euler:
https://archive.org/details/elementsofalgebr00eule

as well as the more complicated method for solving 4th degree equations. Equations of 5th degree and higher cannot be solved in general by formulas involving only the taking of roots, and can only be approxmated by numerical methods such as newton's method.