Consider signs of $b_n$, $c_n$.

$b_n\equiv 0$, $c_n\equiv 0$

Suppose $b_1\not=0$ and $c_1\not=0$. Then, there's three cases. $$\begin{array}{|c|c|c|c|c|} \hline n & 1 & 2 \\ \hline b_n & + & - \\ \hline c_n & + & - \\ \hline \end{array} \quad\text{or}\quad \begin{array}{|c|c|c|c|c|} \hline n & 1 & 2 & 3 \\ \hline b_n & - & + & - \\ \hline c_n & + & + & - \\ \hline \end{array} \quad\text{or}\quad \begin{array}{|c|c|c|c|c|} \hline n & 1 \\ \hline b_n & - \\ \hline c_n & - \\ \hline \end{array}$$For every case, there exists $n$ s.t. $b_n,c_n<0$. Then \begin{align*} b_{n+1}^2 - 4c_{n+1} &= \left(\frac{-b_n - \sqrt{b_n^2 -4c_n}}{2} \right)^2 - 4 \cdot \frac{-b_n + \sqrt{b_n^2 -4c_n}}{2} \\ &= \left( \frac{b_n^2}{2} +2b_n -c_n \right) - \left( \frac{-b_n}{2} + 2\right) \sqrt{b_n^2 -4c_n} \\ &< \left( \frac{b_n^2}{2} +b_n \right) - \left( \frac{-b_n}{2} + 2\right) (-b_n) = 3b_n <0 \end{align*}Hence, $x^2+b_{n+1}x+c_{n+1}=0$ does not have a real root. Contradiction! If $b_1=0$, then $c_1\geq 0$ but at the same time $c_1\leq 0$ for $x^2+b_1 x+c_1=0$ has a real root. Hence $c_1=0$. If $c_1=0$, then $b_2=0$, $c_2=-b_1$. By similar argument as the above case, we get $c_2=0$. Hence $b_1=0$.

The sequence will end at - + if it's reached from a - - pair that is not the starting pair.