Решите уравнение: x^2+19x+48=0.

\[x² + 19x + 48 = 0\ \]

\[D = 361 - 192 = 169\]

\[x_{1,2} = \frac{- 19 \pm 13}{2}\]

\[x_{1} = \frac{- 6}{2} = - 3;\ \ \ \ x_{2} = \frac{32}{2} = 16.\]