# Real Solutions for a Quartic Equation

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.

• Besides brute force mentioned to tackle general quartic equations, you may use factor theorem to find out possible linear factors (by trial-and-error). Another trick is to see whether it's reducible to the form of $A(x^2+ax+b)(x^2+cx+d)$ by equating coefficients. – Ng Chung Tak Aug 3 '16 at 1:03