Find two consecutive positive integers, sum of whose squares is 365

Let the first number be x, then the next number will be x+1. Then, x²+(x+1)²=365
Given ⇒x²+x²+2x+1=365
⇒2x²+2x−364=0
⇒x²+x−182=0
⇒x²+14x−13x−182=0
⇒x(x+14)−13(x+14)=0
⇒(x−13)(x+14)=0
⇒x=13,−14
Considering positive number, the numbers are 13 and 14.