Решение:
\[ \begin{cases} x + y = -20 \\ x \cdot y = 75 \end{cases} \]
\[ t^2 - (-20)t + 75 = 0 \]
\[ t^2 + 20t + 75 = 0 \]
D = b2 - 4ac = 202 - 4 1 75 = 400 - 300 = 100
\[ t = \frac{-b \pm \sqrt{D}}{2a} = \frac{-20 \pm \sqrt{100}}{2 \cdot 1} = \frac{-20 \pm 10}{2} \]
t1 = \frac{-20 + 10}{2} = \frac{-10}{2} = -5
t2 = \frac{-20 - 10}{2} = \frac{-30}{2} = -15
Ответ: -15-5