No, it depends on the sign of the discriminant. If the discriminant is negative, the quadratic has no solution (at least, no real solutions), and so it cannot be further simplified. If the discriminant is zero, it means that the polynomial has a zero #x_0# with molteplicity two, and this means that #p(x)=(x-x_0)^2#. You can fully solve the equation (and by this I mean writing #p(x)=(x-x_1)(x-x_2)#, with #x_1\ne x_2# if and only if the discriminant is strictly positive.