Am I supposed to consider or solve? Let's solve.
Let's call our integers x and x+1
square of the second integer added to 3 times the first is 37
(x+1)² + 3x = 37
x² + 2x + 1 + 3x - 37 = 0
x² + 5x - 36 = 0
(x+9)(x-4) = 0
Answer: x = -9 or x=4
Check:
x=-9
(-9+1)² + 3(-9) = 64 - 27 = 37 good
x=4
(4+1)² + 3(4) = 25 + 12 = 37 good
Two solutions.
