Today's Problem

https://ru.wikipedia.org/wiki/%D0%A4%D0%BE%D1%80%D0%BC%D1%83%D0%BB%D0%B0_%D0%9A%D0%B0%D1%80%D0%B4%D0%B0%D0%BD%D0%BE

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The original foray into complex numbers - by the Renaissance
Italian mathematician Cardano - was in aid of finding the roots
of cubic equations, a daunting practical task at the time.  In effect,
it was impossible to tell if any said equation had positive roots
without applying his formula. This will take some unraveling...
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First, let us recall how to evaluate a higher order sum, with a familiar
formula. It has a name,  the Binomial Theorem:

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Next, we need to look at a modern formula which solves the simplest

higher order equation, the quadratic.

For (ax^2 + bx +c = 0):

The quadratic formula is actually a piece of reverse engineering,
ie from a buit-up construct, going in reverse to find the elements.
René Descartes was the first to use it, but could one derive the logic
behind it from the usual protocol used to solve quadratics?

Let's look at an example.
(x+2)(x+7)=
 x^2 + 9x + 14    which is:
ax^2 + bx + c      with a=1; b=9; and c=14.

What is the first thing I woud do, if asked to solve the equation?

I would ask myself what the possible factorizations of 14
is adequate to balancing the coefficients:
(b^2 - 2^2ac)^.5 = 5

How does it look with respect to b, given the play of signs?
-b + or - (b^2 - 2^2ac)^.5 = -4, -14

a here is 1, so there are two single x creating factors to account for:
-b + or - (b^2 - 2^2ac)^.5 /2a= -2, -7

Solving for zeros:
x - 2 = 0; x = 2
x - 7 = 0; x = 7
2 and 7 are then the roots of the equation.

An actual mathematical derivation of the formula - and
there are many - can be had on Wikipedia.

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If quadratics were easy enough to sole by intuition, or trial
and error, cubic equations were harder. Moreover, the fact
that one might face an equation with no actual roots could
not be ignored indefinitely. Cardan first published on a
systematic approach to some cubics. Bombelli then tackled
the issue of imaginary numbers, and how to use mathematical
operations on them.

Below, a change of constant term shows the funtion crosses
at 2 and -2.