Algebra Question?

 The sum of three consecutive positive integers

 is equal to 40 less than the square of the middle integer.

 Determine the three integers.

 (x - 1) + x + (x + 1) = x^2 - 40

 x^2 - 3x - 40 = 0

 (x - 8)(x + 5) = 0

 The three integers are 7, 8, and 9. 

The sum of three consecutive positive integers:

I'll call them:

(x - 1), x, and (x + 1)

So the sum of the three is:

x - 1 + x + x + 1

3x

Is equal to 40 less than the square of the middle integer:

3x = x² - 40

Now we can solve for x and throw out any negative values:

0 = x² - 3x - 40

0 = (x - 8)(x + 5)

x = -5 and 8

Throwing out the negative value:

x = 8

Your three numbers are:

7, 8, and 9

7 + 8 + 9= 24

The middle number is 8.

The square of 8 is 64. 

24 is 40 less than the square of 8.