Ferrari's Method

Theorem
A polynomial equation of the form $$a x^4 + b x^3 + c x^2 + d x + e = 0$$ is called a quartic equation, or just quartic.

It has solutions $$x = \frac {-p \pm \sqrt {p^2 - 4 q}} {4}$$ where:
 * $$p = \frac b a \pm \sqrt {\frac {b^2} {a^2} - \frac {4 c} {a} + 4 y_1}$$;
 * $$q = y_1 \mp \sqrt {y_1^2 - \frac {4 e} {a}}$$,

where $$y_1$$ is a real solution to the cubic:
 * $$y^3 - \frac c a y^2 + \left({\frac {b d} {a^2} - \frac {4 e} {a}}\right) y + \left({\frac {4 c e} {a^2} - \frac {b^2 e} {a^3} - \frac {d^2} {a^2}}\right) = 0$$

Ferrari's method is a technique for solving this quartic.

Proof
First we render the quartic into monic form:
 * $$x^4 + \frac {b} {a} x^3 + \frac {c} {a} x^2 + \frac {d} {a} x + \frac {e} {a} = 0$$

Completing the square in $$x^2$$:
 * $$\left({x^2 + \frac {b} {2a} x}\right)^2 + \left({\frac {c} {a} - \frac {b^2} {4 a^2}}\right) x^2 + \frac {d} {a} x + \frac {e} {a} = 0$$

Then we introduce a new variable $$y$$:
 * $$\left({x^2 + \frac {b} {2a} x + \frac y 2}\right)^2 + \left({\frac {c} {a} - \frac {b^2} {4 a^2} - y}\right) x^2 + \left({\frac {d} {a} - \frac {b} {2a} y}\right) x + \left({\frac {e} {a} - \frac {y^2} {4}}\right) = 0$$

This equation is valid for any $$y$$, so let us pick a value of $$y$$ so as to make:
 * $$\left({\frac {c} {a} - \frac {b^2} {4 a^2} - y}\right) x^2 + \left({\frac {d} {a} - \frac {b} {2a} y}\right) x + \left({\frac {e} {a} - \frac {y^2} {4}}\right)$$

have a zero discriminant. That is:
 * $$\left({\frac {d} {a} - \frac {b} {2a} y}\right)^2 = 4 \left({\frac {c} {a} - \frac {b^2} {4 a^2} - y}\right) \left({\frac {e} {a} - \frac {y^2} {4}}\right)$$

After some algebra, this can be expressed as a cubic in $$y$$:
 * $$y^3 - \frac c a y^2 + \left({\frac {b d} {a^2} - \frac {4 e} {a}}\right) y + \left({\frac {4 c e} {a^2} - \frac {b^2 e} {a^3} - \frac {d^2} {a^2}}\right) = 0$$

Using (for example) Cardano's Formula, we can find a real solution of this: call it $$y_1$$.

Now a Quadratic Equation $$p x^2 + q x + r$$ can be expressed as:
 * $$p \left({\left({x + \frac {q} {2p}}\right)^2 - \frac {q^2 - 4 p r} {4 p^2}}\right)$$

If that quadratic has a zero discriminant, i.e. $$q^2 = 4 p r$$, then this reduces to:
 * $$p \left({\left({x + \frac {q} {2p}}\right)^2}\right)$$

... which in turn becomes
 * $$p \left({\left({x + \pm \sqrt{\frac {r} {p}}}\right)^2}\right)$$

as $$q^2 = 4 p r \Longrightarrow \frac {q^2} {4 p^2} = \frac r p$$.

So, as
 * $$\left({\frac {c} {a} - \frac {b^2} {4 a^2} - y_1}\right) x^2 + \left({\frac {d} {a} - \frac {b} {2a} y_1}\right) x + \left({\frac {e} {a} - \frac {y_1^2} {4}}\right)$$

has a zero discriminant (we picked $$y_1$$ to make that happen), we can write it as:
 * $$\left({\frac {c} {a} - \frac {b^2} {4 a^2} - y_1}\right)\left({x \pm \frac {\sqrt{\left({\frac {e} {a} - \frac {y_1^2} {4}}\right)}} {\sqrt{\left({\frac {c} {a} - \frac {b^2} {4 a^2} - y_1}\right)}}}\right)^2$$

Now we return to the equation:
 * $$\left({x^2 + \frac {b} {2a} x + \frac {y_1} 2}\right)^2 + \left({\frac {c} {a} - \frac {b^2} {4 a^2} - y_1}\right) x^2 + \left({\frac {d} {a} - \frac {b} {2a} y_1}\right) x + \left({\frac {e} {a} - \frac {y_1^2} {4}}\right) = 0$$

which can now be written:
 * $$\left({x^2 + \frac {b} {2a} x + \frac {y_1} 2}\right)^2 = \left({\frac {b^2} {4 a^2} - \frac {c} {a} + y_1}\right)\left({x \mp \frac {\sqrt{\left({\frac {y_1^2} {4} - \frac {e} {a}}\right)}} {\sqrt{\left({\frac {b^2} {4 a^2} - \frac {c} {a} + y_1}\right)}}}\right)^2$$

Taking square roots of both sides:
 * $$x^2 + \frac {b} {2a} x + \frac {y_1} 2 = \pm x \sqrt {\left({\frac {b^2} {4 a^2} - \frac {c} {a} + y_1}\right)} \mp \sqrt{\frac {y_1^2} {4} - \frac {e} {a}}$$

Arranging into canonical quadratic form:
 * $$x^2 + \left({\frac {b} {2a} \pm \frac 1 2 \sqrt {\frac {b^2} {a^2} - \frac {4 c} {a} + 4 y_1}}\right) x + \frac 1 2 \left({y_1 \mp \sqrt{y_1^2 - \frac {4 e} {a}}}\right) = 0$$

This quadratic in $$x$$ can then be solved in the conventional manner.

Hence the result.