Решите уравнение: (4x-3)^2+(2x-1)(2x+1)=24.

\[20x² - 24x - 16 = 0\ \ |\ :4\]

\[5x^{2} - 6x - 4 = 0\]

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

\[= 36 + 80 = 116\]

\[x_{1} = \frac{6 + \sqrt{116}}{2 \cdot 5} = \frac{6 + 2\sqrt{29}}{10} =\]

\[= \frac{3 + \sqrt{29}}{5}\]

\[x_{2} = \frac{6 - \sqrt{116}}{2 \cdot 5} = \frac{6 - 2\sqrt{29}}{10} =\]

\[= \frac{3 - \sqrt{29}}{5}\]

\[Ответ:\ x = \frac{3 \pm \sqrt{29}}{5}.\]