Question 1107916
.
Let the number be 'x'


The square of the number is x^2 


3 more than the number is (x+3);


The square of 3 more than the number is (x+3)^2


So x^2+(x+3)^2 = 29


x^2+(x^2+6x+9)-29=0


2x^2+6x-20=0


x^2+3x-10=0


(x+5)(x-2)=0


x=-5 and 2


But since x must be a positive number, x=2.


<U>Answer</U>.  The number is 2.