Решите уравнение: (6x-5)^2+(3x-2)(3x+2)=36.

\[45x² - 60x - 15 = 0\ \ |\ :15\]

\[3x^{2} - 4x - 1 = 0\]

\[D = ( - 4)^{2} - 4 \cdot 3 \cdot ( - 1) =\]

\[= 16 + 12 = 28\]

\[x_{1} = \frac{- ( - 4) + \sqrt{28}}{2 \cdot 3} = \frac{4 + 2\sqrt{7}}{6} =\]

\[= \frac{2 + \sqrt{7}}{3}\]

\[x_{2} = \frac{- ( - 4) - \sqrt{28}}{2 \cdot 3} = \frac{4 - 2\sqrt{7}}{6} =\]

\[= \frac{2 - \sqrt{7}}{3}\]

\[Ответ:\ \ x = \frac{2 + \sqrt{7}}{3};\ x = \frac{2 - \sqrt{7}}{3}.\]