Question 658159
Let the two numbers be x and x+1
{{{x^2 + (x+1)^2 = 61}}}
{{{x^2 + (x^2 + 2x + 1) = 61}}}
{{{2x^2 + 2x - 60 = 0}}}
{{{x^2 + x - 30 = 0}}}
We can solve for x using the quadratic equation {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} with a=1, b=1, and c=-30:
{{{x = (-1 +- sqrt( 1^2-4*1*(-30) ))/(2*1) }}}
{{{x = (-1 +- sqrt( 1-(-120) ))/2 }}}
{{{x = (-1 +- sqrt( 121 ))/2 }}}
{{{x = (-1 +- 11)/2 }}}
And since we're only looking for positive integers:
{{{x = (-1 + 11)/2 }}}
{{{x = 10/2 = 5 }}}
So the two numbers are 5 and 6