Because the sum of the infinite geometric series is \(\frac{N}{1-r},\)
and the first term (N) is 8/5*1/r=8/5r, \(10=\frac{\frac{8}{5r}}{1-r}\rightarrow 10-10r=\frac{8}{5r}\rightarrow 50r-50{r}^{2}=8\rightarrow -50{r}^{2}+50r-8=0.\)
So a=-50, b=50, c=-8. And if the discriminant (b squared minus (4 times a times c)) is 0, then there is only one possible solution (including imaginary solutions). And because 50^2-(4*-50*-8) = 900 is not 0, there are two solutions. And using the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
x=r is 0.2 or 0.8.
1st case: r=0.2, N=8/(5*0.2)=8.
2nd case: r=0.8, N=8/(5*0.8)=2.
