x4 + px2 + qx + r = 0
Add px2+p2 to each side, and move qx and r to the right, so that the left side becomes the square of x2+p. Then introduce a new variable y into this square as shown below.
(x2+p)2 = px2 - qx - r + p2
(x2+p+y)2 = (p+2y)x2 - qx - r + p2 + y2 + 2py
This is a polynomial in two variables, x and y. If x was a solution before, it is still a solution, along with every value of y. Instead of four solution points (x), we have four solution lines (x cross y). Any value of y will do.
Adjust y so that the right side becomes a square. In other words, the right side should be (ux+v)2, for some u and v. This happens iff the discriminant on the coefficients is 0. The lead coefficient is p+2y, the linear coefficient is -q, and the constant term is -r+p2+y2+2py. The discriminant b2-4ac becomes a cubic polynomial in y. Solve this using the cubic procedure, and use any of the three solutions for y. Use y to compute u and v, and write
x2+p+y = ±(ux+v)
Solve this by the quadratic formula, giving up to four solutions for x.
This procedure won't work in rings with characteristic 2 or 3.
Should we try one? How about a polynomial with small integer solutions.
x4-25x2+60x-36 = 0
(x2-25)2 = -25x2 -60x + 661
(x2-25+y)2 = (2y-25)x2 -60x + 661 + y2 - 50y
-8y3 + 500y2 - 10288y + 69700 = 0
y = 17 (a solution for y, using the cubic procedure)
x2-8 = ±(3x-10)
x2-3x+2 = 0 and x2+3x-18 = 0
x = {1,2,3,-6}