Analytic expressions: the quartic equation
After the cubic equation we will now consider the quartic (fourth-degree) equation: ax4+bx3+cx2+dx+e=0. We may assume that a=1 and also that b=0. Indeed, because a≠0 we can divide all coefficients by a and after that we can get rid of the (new) b by writing x as y-b/4; after some manipulations we end up with an equation of the form y4+py2+qy+r=0.
If q=0 then we actually have a quadratic equation for y2, which we can solve using the quadratic formula and then solve the original equation by taking square roots.
If q≠0 then we can use Ferrari's method to create two separate quadratic equations for  y. On the way to these equations you encounter a cubic equation, which can be solved using yesterday's method. The resulting formulas look impressive but, and that is important for what comes later, they are built up using addition/subtraction, multiplication/division and roots: square roots and cubic roots (also fourth roots but you get these by taking square roots twice).
This was the state of the art at the end of the sixteenth century and the mathematicians went to work to try to solve the quintic (fifth-degree) equation in the same way as well and that brings us to the `why' of this question: it just would not work. And in the beginning of the nineteenth century they discovered why: there is no formula for the solution of the fifth- and higher-degree equations. How that works we will see tomorrow.
