[Algebra]
Problem: Let $T=$ $\frac{7}{8}$. The number $T$ can be expressed as a reduced fraction $\frac{m}{n}$, where $m$ and $n$ are positive integers whose greatest common divisor is 1 . The equation $x^{2}+(m+n) x+m n=0$ has two distinct real solutions. Compute the lesser of these two solutions.

Solution: The left-hand side of the given equation can be factored as $(x+m)(x+n)$. The two solutions are therefore $-m$ and $-n$, so the answer is $\min \{-m,-n\}$. With $T=\frac{7}{8}, m=7, n=8$, and $\min \{-7,-8\}$ is $\mathbf{- 8}$.

Final answer: -8