I'm not sure what to do about this problem, so please help me out.

21

Step-by-step explanation:

Let the three consecutive integers be;

    x , x + 1 and x + 2:

The sum of squares of the two smaller numbers:

  The two smaller numbers are x and x+ 1

Sum of their squares:

           x² + (x+1)²;

        x² + x² + 2x + 1 = 2x² + 2x + 1

It is 45 more than the square of the larger number:

    Larger number = x + 2

        (x+2)² = x² + 4x + 4

 45 more;

        x² + 4x + 4 + 45  = x² + 4x + 49

Both are equal:

 2x² + 2x + 1 =  x² + 4x + 49  

  2x² - x² + 2x - 4x + 1 - 49 = 0

      x² - 2x - 48 = 0

      x² - 6x + 8x - 48 = 0

      x(x - 6) + 8(x - 6) = 0

      (x + 8)(x-6) = 0

      x + 8 = 0 or x - 6 = 0

      x  = -8 or x = 6

We choose x = 6 because it makes the solution true as a positive integer

  Sum of the three numbers:

     x + x + 1 + x + 2 = 6 + 6 + 1 + 6 + 2 = 21