The quadratic equation has two solutions if b^2 - 4ac > 0
Given the equation ax^4 + bx^2 + c=0
Let P = x^2
Substitute into the formula to have:
a(x^2)^2 + bx^2 + c =0
The equation becomes aP^2 + bP + c = 0
For us to have a unique solution, the discriminant b^2 - 4ac must be greater than zero. Hence the quadratic equation has two solutions if b^2 - 4ac > 0
