Quadratic #2

No way this is a level 4 problem. Overrated

Let the roots of the equation be a,a+1 Therefore equation would be x^2-(2a+1)x+(a(a+1)) Therefore b^2-4ac=4a^2+1+4a-4a^2-4a =1

let n be the first integer:
-b=-n-(n+1)=-2n-1 (from (x-n)[x-(n+1)]);
b=2n+1;
c=n^2+n;
b^2-4c=(2n+1)^2-4(n^2+n)=(4n^2+4n+1)-4n^2-4n=1

Let the first root be $n$ .Then the 2nd root will be $n+1$ .So the equation will be $(x-n)(x-(n+1))=x^2-(n+(n+1))x+n(n+1)=x^2-(2n+1)x+n(n+1)$ . So $b=2n+1,c=n(n+1)\rightarrow b^2-4c=(2n+1)^2-4n(n+1)=(4n^2+4n+1)-(4n^2+4n)=\boxed{1}$