I interpret this as the sum of groups of terms, each group represented by √x2+√(x2+1)+√(x2+2) where x is a natural number. So if we expand the series a little we get:
√1+√2+√3+√4+√5+√6+√9+√10+√11+√16+√17+√18+...+√x2+√(x2+1)+√(x2+2)=5190.
In this interpretation the first member of each group is a perfect square (1, 4, 9, 16, ...). Therefore its square root is (1, 2, 3, 4, ...). The sum up to integer x is x(x+1)/2. The remaining part of the sum will be a little greater than twice this expression and will be irrational. Therefore the sum for the whole series>3x(x+1)/2. For example, the 4th group is √16+√17+√18=4+4.123+4.243=12.366 approx., which is a little greater than 3×4=12. As x increases the discrepancy will decrease. We can do a rough estimate of x by solving:
3x(x+1)/2<5190, that is, x(x+1)<3460, x2+x-3460=0 from which x=59 to the nearest integer.
The sum involves irrational numbers, yet the sum is a rational integer, which seems unlikely.
It would make more sense if the equality was replaced by an inequality.
If the interpretation is correct then it seems expedient to solve the problem using a computer program like the following:
x=0, y=0;
1: x=x+1;
y=y+x+√(x2+1)+√(x2+2);
IF y<5190 THEN GOTO 1;
PRINT x;
When this program is run the result is 59, which corresponds to the rough estimate of the solution of the inequality.
The actual sum = 5316.5314 approx when x=59. When x=58 the sum is 5139.5060 approx.
Experiments and logic show that E=3x(x+1)/2 gives a fairly accurate assessment of the actual sum of the series, even for quite low values of x. For example, if the S≥L, where L is the given limit, and we need to find x when the sum is at least L=50, then x=6 and S=66.22 approx, compared with E=63 (about 95% of the true value). For L=100, x=6, S=111.62 approx and E=108 (97%). For L=5190, x=59, S=5316.53 approx and E=5310 (99.9%).