\[h(x) = \frac{x^{2} + 3}{x - 3};\ \ \ h(x) = 19:\]
\[\frac{x^{2} + 3}{x - 3} = 19;\ \ \ \ \ \ \ x \neq 3\]
\[x^{2} + 3 - 19x + 57 = 0\]
\[x^{2} - 19x + 60 = 0\]
\[x_{1} + x_{2} = 19;\ \ \ \ \ \ \ \ \ \ \ \ x_{1} = 4\]
\[x_{1} \cdot x_{2} = 60;\ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{2} = 15\]
\[Ответ:при\ x = 4;x = 15.\]