Question 281263
Th integers r
{{{n}}} and
{{{n + 2}}}
given:
{{{n^2 + 3*(n + 2) = 24}}}
{{{n^2 + 3n + 6 - 24 = 0}}}
{{{n^2 + 3n - 18 = 0}}}
{{{(n + 6)*(n - 3) = 0}}}
{{{n = -6}}} (can't be minus)
{{{n = 3}}}
{{{n + 2 = 5}}}
The numbers are 3 and 5
check:
{{{n^2 + 3*(n + 2) = 24}}}
{{{3^2 + 3*5 = 24}}}
{{{9 + 15 = 24}}}
{{{24 = 24}}}
OK