Question 351681

 Let the number is x.

 now according to question as 5 more than the square of a number ( i.e 5 + x*x )
is 6 times the number ( i.e 6x )

so, 5 + x^2 = 6x
 
 =>  x^2 + 5 -6x =0 

 =>  x^2 - 6x +5 = 0

 =>  x^2 -5x -x + 5 = 0    
                     (here we factorize 5 such that it become -6 after addition)

now x(x -5) -1 (x -5) =0
 
=>  (x-5) (x -1) = 0

so, x-5 = 0   or        x-1 =0
     x = 5              x = 1


so, required numbers are 1 and 5.



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