= Solution
For a nonidentity element of $\operatorname{PSL}_2(\mathbb R)$ represented by a determinant-one matrix, the classification is elliptic, parabolic or hyperbolic according as the squared <trace> is less than, equal to or greater than four. This follows either from the <eigenvalues> or from the <discriminant> of its fixed-point equation on the boundary of the <upper half-plane>.
The first matrix has <determinant> one and <trace> two, and is not scalar. It is therefore \b[parabolic]. Its fixed-point equation is
$$
z=\frac{5z+8}{-2z-3}\iff 2(z+2)^2=0.
$$
Thus \b[its unique <fixed point> is $z=-2$ on the real boundary], of multiplicity two; infinity is not fixed and there is no <fixed point> in the <upper half-plane>. The second matrix has <determinant> one and <trace> $-4$, so it is \b[hyperbolic]. Its two boundary <fixed points> are $( -1\pm\sqrt3)/2$, consistently with this classification.
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