I am looking for the real solutions for the equation: $$Ax^4+Bx^3+Cx^2+Dx+E=0$$ I followed the wikipedia page on quartics and found the discriminant, P, and Q such that the cases of solutions are:
If ∆ < 0 then the equation has two distinct real roots and two complex conjugate non-real roots.
If ∆ > 0 then either the equation's four roots are all real or none is.
If P < 0 and D < 0 then all four roots are real and distinct.
If P > 0 or D > 0 then there are two pairs of non-real complex conjugate roots. If ∆ = 0 then (and only then) the polynomial has a multiple root. Here are the different cases that can occur: If P < 0 and D < 0 and ∆0 ≠ 0, there is a real double root and two real simple roots.
If D > 0 or (P > 0 and (D ≠ 0 or Q ≠ 0)), there is a real double root and two complex conjugate roots.
If ∆0 = 0 and D ≠ 0, there is a triple root and a simple root, all real.
If D = 0, then:
If P < 0, there are two real double roots.
If P > 0 and Q = 0, there are two complex conjugate double roots.
If ∆0 = 0, all four roots are equal to $-\frac{b}{4a}$
Where $\Delta$, P, Q, D, $\Delta_0$ are as defined in https://en.wikipedia.org/wiki/Quartic_function
I was curios if someone could point me in the correct direction as I was very confused by the rest of the Wikipedia page.
Once again, I only need the real solutions.
Thanks in advance, Zach Hilman