Question 962495
Let a be the larger integer and b be the smaller integer. We know that
a = b + 2
since they're consecutive even integers and a is the larger one. Also,
{{{a^2-10b=76}}}
as we are told. Substituting, we get
{{{(b+2)^2-10b=76}}}
{{{b^2+4b+4-10b=76}}}
{{{b^2-6b+4=76}}}
{{{b^2-6b-72=0}}}
We use the quadratic formula to get
{{{b = (-(-6) +- sqrt( (-6)^2-4*1*-72 ))/(2*1) }}}
{{{b=(6+-sqrt(324))/2}}}
{{{b=(6 +- 18)/2}}}
{{{b=3 +- 9}}}
So b could be 12 or -6, but since the question says the numbers are positive, b must be 12. We conclude that a is 14.

Check:
{{{14^2-10(12)=196-120=76}}}