If roots of the equation $x^2-bx+c=0$ are two consecutive integers, then $b^2-4c$ equals:

Let the roots be $\alpha$ and $\alpha +1$.Sum of roots: $\alpha +(\alpha +1)=b⟹2\alpha +1=b$ Product of roots: $\alpha (\alpha +1)=c⟹{\alpha }^2+\alpha =c$ We need to find the value of the discriminant $D=b^2-4c$:
$$b^2-4c=(2\alpha +1{)}^2-4({\alpha }^2+\alpha )$$
$$b^2-4c=(4{\alpha }^2+4\alpha +1)-(4{\alpha }^2+4\alpha )$$
$$b^2-4c=1$$