Is there a quadratic equation for matrices? If you

Yes. If a matrix satisfies a polynomial, it implies that all its eigenvalues satisfy this same polynomial. (For example in your situation, if v is an eigenvector of A with eigenvalue ${\displaystyle \lambda }$, then ${\displaystyle a{A}^{2}v+bAv+c{I}_{n}v=(a{\lambda }^2+b\lambda +c)v=0}$, and since v is nonzero this implies that ${\displaystyle a{\lambda }^2+b\lambda +c=0}$.
In particular, you know that A has at most two eigenvalues, and in particular they could be either of the roots of the polynomial ${\displaystyle a{X}^2+bX+c}$. (If it happens that this polynomial has only one root, then you know A has only one eigenvalue.)
Using the Jordan form, it tells you that A's Jordan form has only at most these two values on the diagonal; thus A is similar to a Jordan form matrix with at most those two values on the diagonal.
Actually, you can know a little more. If the quadratic polynomial ${\displaystyle a{X}^2+bX+c}$ has two distinct roots, then the matrix must be diagonalizable. Indeed, a Jordan block of size ${\displaystyle m\times m}$ with eigenvalue α has minimal polynomial ${\displaystyle {(X-\alpha )}^m}$, and your matrix can't satisfy a polynomial one of its Jordan blocks doesn't satisfy, so if your polynomial has the form ${\displaystyle a(X-\alpha )(X-\beta )}$ with ${\displaystyle \alpha \ne \beta }$, then it cannot have a jordan block of size ${\displaystyle m>1}$. So all the Jordan blocks are size 1. To summarize, if your polynomial has 2 distinct roots, then A is similar to a diagonal matrix with at most those 2 values on the diagonal.
If your polynomial has a double root, i.e., it's of the form ${\displaystyle a{(X-\alpha )}^2}$, then its Jordan form can have blocks of size up to 2. So in this case, it's similar to a Jordan matrix with only one value on the diagonal and blocks of size 1 and/or 2.
This is the most you can say though. Any matrix like I have described will be satisfy your polynomial, because all its Jordan blocks do.