\[10b^{2} - 7b + 1 = 0\]
\[D = 49 - 40 = 9\]
\[b_{1} = \frac{7 + 3}{20} = \frac{10}{20} = 0,5;\ \ \]
\[b_{2} = \frac{7 - 3}{20} = \frac{4}{20} = \frac{1}{5} = 0,2.\]