This is about Project Euler Problem 27. The question is:
Considering quadratics of the form $n^2 + an + b$, where $\lvert a \rvert < 1000$ and $\lvert b \rvert < 1000$
Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting with $n = 0$.
I've solved the problem using a search through the values of $a$ and $b$, and have improved it by reducing the search space considerably, but I don't understand how the following solution works:
problem_27 = -(2*a-1)*(a^2-a+41)
where n = 1000
m = head $ filter (\x->x^2-x+41>n) [1..]
a = m-1
Pseudocode
Let n = 1000
Let m = The first i in [1,2..] such that i^2 - i + 41 > n
Let a = m - 1
Then the solution is -(2a - 1)(a^2 - a + 41)