Решите уравнение: (4x-1)^2-4(x-2)(x+2)=0.

\[(4x - 1)^{2} - 4(x - 2)(x + 2) = 0\]

\[16x^{2} - 8x + 1 - 4\left( x^{2} - 4 \right) = 0\]

\[16x^{2} - 8x + 1 - 4x^{2} + 16 = 0\]

\[12x^{2} - 8x + 17 = 0\]

\[D_{1} = 16 - 204 < 0\]

\[нет\ корней.\]