\(D = b^2 - 4ac = (-3)^2 - 4 \cdot 4 \cdot (-1) = 9 + 16 = 25\)
\(x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + 5}{2 \cdot 4} = \frac{8}{8} = 1\)
\(x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - 5}{2 \cdot 4} = \frac{-2}{8} = -\frac{1}{4}\)
Ответ: $$x_1=1, x_2=-\frac{1}{4}$$