Question 1160519
use units of gallon-deg
have 30000*78=2,340,000 gal-deg now
want to have  (30000+x)*80=2,400,000+80x gal-deg
and the initial amount is increased by 90x gal-deg, where x is the number of gallons of 90 degree water.
so 2340000+90x=2400000+80x
10x=60000
x=6000 gallons

Another way is that there is a 2 degree deficit for 30K gallons, or 60K deficit of gal-deg
you are adding water that is a 10 degree surplus over what you want, and 6000 gallons of that is 60K gal-deg.