Let us consider a graph $G$ having $m$ number of complete sub-graphs $K\_{n\_1},K\_{n\_2},...,K\_{n\_m}$ which have size $n\_1,n\_2,...,n\_m$ respectively. Further $\forall i$, one vertex of $K\_{n\_i}$ is connected to one vertex of $K\_{n\_{i+1}}$ by an edge. Similarly, one vertex (different from previous one) of $K\_{n\_i}$ is connected to one vertex of $K\_{n\_{i-1}}$ by an edge. In this way, complete sub-graphs are connected in chain to form $G$. Find the eigenvalues of adjacency matrix of $G$.



Spectrum of adjacency matrix of block graph
The characteristic polynomial of a complete graph with $n$ vertices is $(x-n+1)(x+1)^{n-1}$. By deleting a vertex from the complete graph $K\_n$, the remaining graph is the complete graph $K\_{n-1}$ with characteristic polynomial $(x-n+2)(x+1)^{n-2}$.
Consider subgraph $G\_i$ of the graph $G$ that is constructed from the union of two complete subgraphs $K\_{n\_i}$ and $K\_{n\_{i+1}}$ of $G$ and the edge which connect these two subgraphs. By using the following theorem of Schwenk, you can find the characteristic polynomial of $G\_i$.  

Schwenk A. J., Computing the characteristic polynomial of a graph, in Graphs 
and Combinatorics (eds. Bari, R., Harary, F.), Springer-Verlag (New York), 1974. 



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> > For any edge $uv$ of the graph $G$,  
> > 
> >  \begin{align\*}
> > P\_G(x)=P\_{G-uv}(x)-P\_{G-u-v}(x)-2\sum\_{Z\in C(uv)}P\_{G-V(Z)}(x),
> > \end{align\*}
> >  where $C(uv)$ denotes the set of all cycles containing $uv$.
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> > 
> > 
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Since the edge which connect the two complete subgraphs in $G\_i$ is not contained in any cycles, the third part of the above formula is zero. So, the characteristic polynomial of $G\_i$ is
\begin{align\*}
(x-n\_i+1)(x+1)^{n\_i-1}(x-n\_{i+1}+1)(x+1)^{n\_{i+1}-1}\\
-(x-n\_i+2)(x+1)^{n\_i-2}(x-n\_{i+1}+2)(x+1)^{n\_{i+1}-2}
\end{align\*}
Such as above and by using the above theorem repeatedly, you can find the characteristic polynomial of $G$. 
As Meyerowitz mentioned, you can find some roots of this characteristic polynomial that is equal to $-1$ simply, but finding some of them is difficult in general. 


