Ferrari's Method

Theorem
Let $P$ be the quartic equation:
 * $a x^4 + b x^3 + c x^2 + d x + e = 0$

such that $a \ne 0$.

Then $P$ has solutions:
 * $x = \dfrac {-p \pm \sqrt {p^2 - 8 q} } 4$

where:

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

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

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

Completing the Square in $x^2$:
 * $\paren {x^2 + \dfrac b {2 a} x}^2 + \paren {\dfrac c a - \dfrac {b^2} {4 a^2} } x^2 + \dfrac d a x + \dfrac e a = 0$

Then we introduce a new variable $y$:
 * $\paren {x^2 + \dfrac b {2 a} x + \dfrac y 2}^2 + \paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y} x^2 + \paren {\dfrac d a - \dfrac b {2 a} y} x + \paren {\dfrac e a - \dfrac {y^2} 4} = 0$

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

have a zero discriminant.

That is:
 * $\paren {\dfrac d a - \dfrac b {2 a} y}^2 = 4 \paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y} \paren {\dfrac e a - \dfrac {y^2} 4}$

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

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:
 * $p \paren {\paren {x + \dfrac q {2 p} }^2 - \dfrac {q^2 - 4 p r} {4 p^2} }$

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

which in turn becomes:
 * $p \paren {\paren {x + \pm \sqrt {\dfrac r p} }^2}$

as $q^2 = 4 p r \implies \dfrac {q^2} {4 p^2} = \dfrac r p$.

So, as:
 * $\paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y_1} x^2 + \paren {\dfrac d a - \dfrac b {2 a} y_1} x + \paren {\dfrac e a - \dfrac { {y_1}^2} 4}$

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

Now we return to the equation:
 * $\paren {x^2 + \dfrac b {2 a} x + \dfrac {y_1} 2}^2 + \paren {\dfrac c a - \dfrac {b^2} {4 a^2} - y_1} x^2 + \paren {\dfrac d a - \dfrac b {2 a} y_1} x + \paren {\dfrac e a - \dfrac { {y_1}^2} 4} = 0$

which can now be written:
 * $\paren {x^2 + \dfrac b {2 a} x + \dfrac {y_1} 2}^2 = \paren {\dfrac {b^2} {4 a^2} - \dfrac c a + y_1} \paren {x \mp \dfrac {\sqrt {\paren {\dfrac { {y_1}^2} 4 - \dfrac e a} } } {\sqrt {\paren {\dfrac {b^2} {4 a^2} - \dfrac c a + y_1} } } }^2$

Taking square roots of both sides:
 * $x^2 + \dfrac b {2 a} x + \dfrac {y_1} 2 = \pm x \sqrt {\paren {\dfrac {b^2} {4 a^2} - \dfrac c a + y_1} } \mp \sqrt {\dfrac { {y_1}^2} 4 - \dfrac e a}$

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

Let:
 * $p = \dfrac b a \pm \sqrt {\dfrac {b^2} {a^2} - \dfrac {4 c} a + 4 y_1}$
 * $q = y_1 \mp \sqrt { {y_1}^2 - \dfrac {4 e} a}$

Then equation $(1)$ can be written as:
 * $x^2 + \dfrac p 2 x + \dfrac q 2 = 0$

Using the Quadratic Formula, putting $a = 1, b = \dfrac p 2, c = \dfrac q 2$:

Hence the result.

Also see

 * Cardano's Formula