Question 165493
Let n be the first integer, (n+2) would be the next consecutive odd integer.
{{{n^2+(n+2)^2=290}}}
{{{n^2+(n^2+4n+4)=290}}}
{{{2n^2+4n-286=0}}}
{{{n^2+2n-143=0}}}
You can factor this quadratic equation. 
{{{(n+13)(n-11)=0}}}
.
.
.
First solution:
{{{n+13=0}}}
{{{n=-13}}}
The two integers are then -13 and -11.
.
.
.
Second solution:
{{{n-11=0}}}
{{{n=11}}}
The two integers are then 11 and 13.