$a^2x^2 + 4ax - 5 = 0$
$x = \frac{1}{4}$:
$a^2 * (\frac{1}{4})^2 + 4a * \frac{1}{4} - 5 = 0$
$\frac{1}{16}a^2 + a - 5 = 0$
$D = b^2 - 4ac = 1^2 - 4 * \frac{1}{16} * (-5) = 1 + \frac{5}{4} = 1 + 1,25 = 2,25 > 0$
$a_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{2,25}}{2 * \frac{1}{16}} = \frac{-1 + 1,5}{\frac{1}{8}} = 0,5 * 8 = 4$
$a_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{2,25}}{2 * \frac{1}{16}} = \frac{-1 - 1,5}{\frac{1}{8}} = -2,5 * 8 = -20$
Ответ: при a = −20 и a = 4