Is the Cubic Formula More Complex Than the Quadratic Formula?

[Char. Limit]

The proof of the quadratic formula was so simple, I moved to the proof of the cubic formula with supreme confidence. And found myself awash in as and cs and cubic roots.

Can you turn this equation into a cubic?

x=-\frac{b}{3a}

-\frac{1}{3a}\sqrt[3]{\frac{1}{2}(2b^3-9abc+27(a^2)d+\sqrt{(2b^3-9abc+27(a^2)d)^2-4(b^2-3ac)^3}}

-\frac{1}{3a}\sqrt[3]{\frac{1}{2}(2b^3-9abc+27(a^2)d-\sqrt{(2b^3-9abc+27(a^2)d)^2-4(b^2-3ac)^3}}

 

[rochfor1]

Googling "proof of cubic forumla" gives this proof.

 

[Strafespar]

First, the cubic equation: ax^3+bx^2+cx+d. With x=y-\frac{a}{3}, you can reduce the equation to y^3+py+q. p=b-\frac{a^2}{3} and q=c-\frac{ab}{3}+\frac{2a^3}{27}. In a cubic equation there are 3 possible answers, the one you listed would be one of the 3, X_{1}