Решите уравнение: 2x^2-4x-17=0.

\[2x² - 4x - 17 = 0\]

\[D = ( - 4)^{2} - 4 \cdot 2 \cdot ( - 17) =\]

\[= 16 + 136 = 152\]

\[x_{1} = \frac{4 + \sqrt{152}}{2 \cdot 2} = \frac{4 + 2\sqrt{38}}{4} =\]

\[= \frac{2 + \sqrt{38}}{2}\]

\[x_{2} = \frac{4 - \sqrt{152}}{2 \cdot 2} = \frac{4 - 2\sqrt{38}}{4} =\]

\[= \frac{2 - \sqrt{38}}{2}\]

\[Ответ:\ x = \frac{2 + \sqrt{38}}{2};\ \ \]

\[x = \frac{2 - \sqrt{38}}{2}.\]