Решите уравнение: корень из (x-3)=19-5x.

\[\sqrt{x - 3} = 19 - 5x\]

\[t = \sqrt{x - 3}\]

\[5t^{2} + t - 4 = 0\]

\[t_{1} = - 1;\ \ \ t_{2} = \frac{4}{5}\]

\[1)\ \sqrt{x - 3} = - 1;\ \ \ \ \]

\[\sqrt{x - 3} \geq 0 \Longrightarrow нет\ решения.\]

\[2)\ \sqrt{x - 3} = \frac{4}{5}\ \]

\[x - 3 = \frac{16}{25}\]

\[x = 3\frac{16}{25}\]

\[Ответ:3\frac{16}{25}\text{.\ }\]