Решите уравнение: (3x-5)^2+(4x-1)(4x+1)=29.

\[25x² - 30x - 5 = 0\ \ |\ :5\]

\[5x^{2} - 6x - 1 = 0\]

\[D = ( - 6)^{2} - 4 \cdot 5 \cdot ( - 1) =\]

\[= 36 + 20 = 56\]

\[x_{1} = \frac{6 + \sqrt{56}}{2 \cdot 5} = \frac{6 + 2\sqrt{14}}{10} =\]

\[= \frac{3 + \sqrt{14}}{5}\]

\[x_{2} = \frac{6 - \sqrt{56}}{2 \cdot 5} = \frac{6 - 2\sqrt{14}}{10} =\]

\[= \frac{3 - \sqrt{14}}{5}\]

\[Ответ:x = \frac{3 + \sqrt{14}}{5};\ \ \]

\[x = \frac{3 - \sqrt{14}}{5}\text{.\ }\]