Question

the sum of two consecutive natural numbers and the square of the first is 5329 what are the numbers

Answer 1

Let The no.s be X,X+1 ; X belongs to Natural no.s
SO,

X+(X+1)+X^2=5329
X^2+2X+1=5329
X^2+2X-5328=0
D = B^2-4AC = 2-(-21312) = 21314
X = [-1 (+or-) √21314]/2 = -2(+or-)146/2{Approx}
X = -1(+or-)73 = -74 or 72

Therefore, "X = 72" (Since, "X" Belongs to Natural no.s)...

Therefore, Required no.s are "72 and 73"

Verification,
72+73+72^2=5329....

HENCE PROVED...