Решите уравнение: 32x^2-12x+1=0.

\[32x^{2} - 12x + 1 = 0\]

\[D = ( - 12)^{2} - 4 \cdot 32 \cdot 1 =\]

\[= 144 - 128 = 16;\ \ \ \ \ \sqrt{D} = 4.\]

\[x_{1} = \frac{12 + 4}{2 \cdot 32} = \frac{16}{64} = \frac{1}{4};\ \ \ \ \ \ \ \ \ \]

\[x_{2} = \frac{12 - 4}{2 \cdot 32} = \frac{8}{64} = \frac{1}{8}\]

\[Ответ:\ \ \frac{1}{4};\ \frac{1}{8}.\]