Solve Any Quartic Equation
==========================
What? In plain text?
Yes, I'm not going to present Ferrari's formula for quartic equation, just
to show you how to do the factorization.

A quartic equation is an equation of form:

  ax⁴+bx³+cx²+dx+e=0

where a is non-zero. So, WLOG let a=1:

  x⁴+bx³+cx²+dx+e=0

Now, let us move  cx²+dx+e to the right-hand side:

       x⁴+bx³=-cx²-dx-e ==>
  ==>  x²(x²+bx)=-cx²-dx-e

The next step will be to make the left-hand side a perfect square, and
I only want to work here with integers, letters and superscript digits,
so let us multiply both sides by 4:

      x²(4x²+4bx)=-4cx²-4dx-4e ==>
  ==> x²[(2x)²+2·2x·b]=-4cx²-4dx-4e

Let us complete the square in the parentheses of the left-hand side, and 
remember to add the same quantity in both sides:

      x²[(2x)²+2·2x·b+b²] = b²x²-4cx²-4dx-4e ==>
  ==> x²(2x+b)² = (b²-4c)x²-4dx-4e ==>
  ==> (2x²+bx)² = (b²-4c)x²-4dx-4e

Now, the left-hand side is a perfect square, is the right-hand side? Not sure!
Let us use the identity

  (x+y)²=x²+2xy+y²

and use some missing variable k to help us make both sides perfect squares:

      (2x²+bx)²+2k(2x²+bx)+k² = (b²-4c)x²-4dx-4e+2k(2x²+bx)+k² ==>
  ==> (2x²+bx+k)² = (b²+4k-4c)x²+(2bk-4d)x+k²-4e

Now, for RHS to be a perfect square, the discriminant should be zero:

  (2bk-4d)-4(k²-4e)(b²+4k-4c)=0

That's a cubic equation in k, and if b,c,d,e are real there is a real k
such that the discriminant is zero. It can be found using the cubic formula.
And the quartic equation can thus be solved using a difference of two squares

Let us try to solve

  x⁴+4x³+6x²+4x+1=0

Here b=4, c=6, d=4, e=1

So,

  (2x²+4x+k)² = (4k-8)x²+(8k-16)x+k²-4

Let us find k:

      (8k-16)²-4(k²-4)(4k-8)=0 ==>
  ==> 64k²-256k+256-4(4k³-8k²-16k+32)=0 ==>
  ==> 64k²-256k+256-16k³+32k²+64k-128=0 ==> 
  ==> 16k³-96k²+192k-128=0 ==>
  ==> k³-6k²+12k-8=0 ==>
  ==> (k-2)³=0 ==>
  ==> k=2

Plug it into the equation:

      (2x²+4x+2)²=0 ==>
  ==> (x²+2x+1)²=0 ==>
  ==> [(x+1)²]²=0 ==>
  ==> (x+1)⁴=0 ==>
  ==> x=-1

The good news is that there is no formulae to solve equations of higher
degree!