Question

the product of two consecutive integers n and n+1 is 42. what is the positive integer that satisfies the situation?

Answer 1

$n(n+1)=42 \\ n^2+n=42 \\ n^2+n-42=0 \\ \\ a=1 \\ b=1 \\ c=-42 \\ b^2-4ac=1^2-4 \times 1 \times (-42)=1+168=169 \\ \\ n=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm \sqrt{169}}{2 \times 1}=\frac{-1 \pm 13}{2} \\ n=\frac{-1 -13}{2} \ \lor \ n=\frac{-1+13}{2} \\ n=\frac{-14}{2} \ \lor \ n=\frac{12}{2} \\ n=-7 \ \lor \ n=6$

The positive integer is 6.
The consecutive integers are 6 and 7.