Question

x^2+(1-5a)x+4a^2-a=0

Answer 1

$x^2+(1-5a)x+4a^2-a=0 \\ D=(1-5a)^2-4\cdot(4a^2-a)=9a^2-6a+1=(1-3a)^2 \\ x_1= \frac{5a-1+1-3a}{2} = a \\ x_2= \frac{5a-1-1+3a}{2}=4a-1$