Given: $r \S t = \frac{(r-t)^2+ rt}{t}$

So, we get:
QUANTITY A: $71 \S 37 = \frac{(71-37)^2+ (71)(37)}{37}=\frac{(34)^2+ (71)(37)}{37}$

QUANTITY B: $37 \S 71 = \frac{(37-71)^2+ (37)(71)}{71}=\frac{(-34)^2+ (37)(71)}{71}$

If we recognize that $(34)^2 = (-34)^2$ and $(71)(37)=(37)(71)$, we can see that both quantities have the exact same numerator

Since the denominator in Quantity A is less than the denominator in Quantity B, we can conclude that Quantity A > Quantity B

Answer: A

Cheers,
Brent