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