Найдите область определения функции: корень из (-2x^2+5x+2).

Ответ:

\[\sqrt{- 2x^{2} + 5x + 2}\]

\[- 2x^{2} + 5x + 2 \geq 0\]

\[2x^{2} - 5x - 2 \leq 0\]

\[D = 25 + 16 = 41\]

\[x_{1,2} = \frac{5 \pm \sqrt{41}}{4};\]

\[\left( x + \frac{5 - \sqrt{41}}{4} \right)\left( x - \frac{5 + \sqrt{41}}{4} \right) \leq 0\]

\[\frac{5 - \sqrt{41}}{4} \leq x \leq \frac{5 + \sqrt{41}}{4}.\]

\[Ответ:\ \frac{5 - \sqrt{41}}{4} \leq x \leq \frac{5 + \sqrt{41}}{4}.\]