Hi Kyla S.,
A fairly interesting problem.
There's a portion which is just slogging through set-up, and then a little bit requiring some reasoning!
If you set up the quadratic formula for solutions for this equation (collect the x^1 terms!) (which you should know by now!) you will note that the square-root term in the numerator must take on integer values, for the term inside of it to be an integer square: 1, 4, 9, 16, etc. (then the square root can be an integer, and the whole expression can aim for integer value (Caution: even though the 2a in the denominator must be checked in a final step). Add in the 28 (=4ac in the quadratic solution), and you are now looking at (5a+1)^2 possibilities of 29, 32, 37, 44, 53, 64, .... Solve these through to (5a+1), and you have the series: √29, 4√2, √37, 2√11, √53, 8, ....
Pretty quickly you might conclude that the "8" is the only promising prospect, but you might have to do some fancier reasoning to prove that this is so (I think it is, because 28+n^2 will only be another perfect square for that single n). Also, you should reason through to show that non-integer values for the (b^2-4ac)^0.5 term would not combine with the (-b) term to end up integer.
That generates only 8 or -8 for possible (5a+1) values; I leave it to you to check these through, including per the Caution: above.
You should take the time to ponder and absorb how this problem was reasoned through, since you are likely to see more problems requiring the same type of analysis.
-- Cheers, -- Mr. d.