How many terms are in an arithmetic series whose sum is 5.586 when t1 = 10 and d = 6?

Sn = n / 2 (2a1 + (n - 1) d)

5586 = n / 2 (20 + (n-1) 6)

5586 = n / 2 (26 - 6n)

5586 = 13n - 3n^2

3n^2 - 13n + 5586 = 0

But I'm confused on how to factor this out since the third term is a large number.

3 answers

you have an error after your second line
5586 = n / 2 (20 + (n-1) 6) ..... ok, but in your sum you had 5.586
5586 = n / 2 (20 + 6n - 6)
5586 = n / 2 (14 + 6n)
5586 = n(7 + 3n)
3n^2 + 7n - 5586 = 0

I used the formula to get 42 and a negative
3n^2 + 7n - 5586 = 0

In order to get the number of terms, wouldn't you have to factor out this equation above

Can I use the discriminant formula?
yes, and it would give you 49 - (4)(3)(-5586) or 67081,
whose square root is 259, so the equation does factor.
However, I feel that having done about 2/3 of the work using the formula, I might as well finish it by the formula
x = (-7 ± 259)/6 = 42 or -133/3

sure enough you equation factors to
(x-42)(3x+133) = 0