Решите уравнение: 2x^2-3x*корень из 6+6=0.

\[2x² - 3x\sqrt{6} + 6 = 0\]

\[D = \left( - 3\sqrt{6} \right)^{2} - 4 \cdot 2 \cdot 6 =\]

\[= 9 \cdot 6 - 48 = 54 - 48 = 6\]

\[x_{1} = \frac{3\sqrt{6} + \sqrt{6}}{2 \cdot 2} = \frac{4\sqrt{6}}{4} = \sqrt{6}\]

\[x_{2} = \frac{3\sqrt{6} - \sqrt{6}}{2 \cdot 2} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}\]

\[Ответ:x = \sqrt{6};x = \frac{\sqrt{6}}{2}.\]