Question

The square of a positive number decreased by twice the number is 48. Find the number

Answer 1

X^2=2x+48
x^2-2x-48=0
(x-8)(x+6)=0
x=8 x=-6
x must be negative so x=-6
check 
36=48+-12
36=36

Answer 2

Let the required positive number is x.

The square of the number is $x^2$

Now, we have been given that square of a positive number decreased by twice the number. It means we have

$x^2-2x$

Now, it is equal to 48. Hence, we have

$x^2-2x=48\\ x^2-2x-48=0\\ x^2-8x+6x-48=0\\ x(x-8)+6(x-8)\\ (x-8)(x+6)=0\\ x=8,-6$

The number is positive, Hence, we have $x=8$

Therefore, the required number is 8