How to solve a first (1st) degree equation

The puzzle:

	a*x + b = 0

Calculate x using the following algorithm:

	x = -b/a

I think I deserve a nobelprize/fieldmedal for that

How to solve a second (2nd) degree equation

The equation

	a*x^2 + b*x + c = 0

has as solutions:

	x = ( -b + sqrt(b^2-4*a*c) ) / (2*a)

	x = ( -b - sqrt(b^2-4*a*c) ) / (2*a)

The equation

	a*x^3 + b*x^2 + c*x + d = 0

first substitute x = y / a^(1/3) , resulting in

	y^3 + aa*y^2 + bb*y + cc = 0

then substitue y = z - aa/3 , resulting in

	z^3 + p*z + q = 0

	with p = bb - aa^2/3   and q = 2*aa^3/27 -bb*aa/3

then suppose z=r+s

	r^3 + 3*r*s(r+s) + s^3 + p*z + q =0

from which follows

	r^3 + s^3 + q = 0
and
	3*r*s + p = 0   (or  27*r^3*s^3 + p^3 = 0 )

from which follows:

	r^3, s^3 = -q/2 {+-} sqrt( (q/2)^2 + (p/3)^3 )

and

	x = (r+s) * exp(2*k*pi*i/3)

... still need to write this down ...