Answer :
[tex]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[/tex]
The positive integer is 6.
The consecutive integers are 6 and 7.
The positive integer is 6.
The consecutive integers are 6 and 7.
