Given a quartic equation $x^4 + ax^3 + bx^2 + cx + d = 0$ apply the Tchirnhaus transformation $x \mapsto y - \frac{a}{4}$ to obtain \begin{equation}\label{reduced} y^4 + py^2 + qy + r = 0 \end{equation}where \begin{eqnarray*} p & = & b - \frac{3a^2}{8}\\ q & = & c - \frac{ab}{2} + \frac{a^3}{8}\\ r & = & d - \frac{ac}{4} + \frac{a^2 b}{16} - \frac{3a^4}{256} \end{eqnarray*}Clearly a solution to Equation () solves the original, so we replace the original equation with Equation (). Move $qy+r$ to the other side and complete the square on the left to get: $$ (y^2+p)^2 = py^2 - qy + (p^2 - r). $$ We now wish to add the quantity $(y^2+p+z)^2 - (y^2+p)^2$ to both sides, for some unspecified value of $z$ whose purpose will be made clear in what follows. Note that $(y^2+p+z)^2 - (y^2+p)^2$ is a quadratic in $y$ Carrying out this addition, we get \begin{equation}\label{middle} (y^2 + p + z)^2 = (p+2z) y^2 - qy + (z^2 + 2pz + p^2 - r) \end{equation}The goal is now to choose a value for $z$ which makes the right hand side of Equation () a perfect square. The right hand side is a quadratic polynomial in $y$ whose discriminant is $$ -8 z^3 - 20pz^2 + (8r-16p^2)z + q^2+4pr-4p^3. $$ Our goal will be achieved if we can find a value for $z$ which makes this discriminant zero. But the above polynomial is a cubic polynomial in $z$ so its roots can be found using the cubic formula. Choosing then such a value for $z$ we may rewrite Equation () as $$ (y^2 + p + z)^2 = (sy + t)^2 $$ for some (complicated!) values $s$ and $t$ and then taking the square root of both sides and solving the resulting quadratic equation in $y$ provides a root of Equation ().