a² + b² = c²

  So substitute (x + 3) for a, (x + 10) for b, and (x + 11) for c

            (x + 3)² + (x + 10)² = (x + 11)²

(x + 3)² = (x + 3)(x + 3) = x² + 3x + 3x + 9 = (x² + 6x + 9)
So substitute (x² + 6x + 9) for (x + 3)²

                   (x² + 6x + 9) + (x + 10)² = (x + 11)²

(x + 10)² = (x + 10)(x + 10) = x² + 10x + 10x + 100 = (x² + 20x + 100)
So substitute (x² + 20x + 100) for (x + 10)²

                  (x² + 6x + 9) + (x² + 20x + 100)² = (x + 11)²

(x + 11)² = (x + 11)(x + 11) = x² + 11x + 11x + 121 = (x² + 22x + 121)
So substitute (x² + 22x + 121) for (x + 11)²

                   (x² + 6x + 9) + (x² + 20x + 100) = (x² + 22x + 121)

                       x² + 6x + 9 + x² + 20x + 100 = x² + 22x + 121

                                    2x² + 26x + 109 = x² + 22x + 121

                                       x² + 4x - 12 = 0

                                     (x + 6)(x - 2) = 0

                                     x + 6 = 0 ;   x - 2 = 0
                                         x = -6;       x = 2

Using x = -6,

a = x + 3 = -6 + 3 = -3  That's no good because we can't have a negative
number for the measure of a side of a triangle.  So we discard -6, 

[not because x is negative but because it produces a negative side of a
triangle.  In other problems like this a negative value for x would result
in only positive sides]  

Using x = 2,

a = x + 3 = 2 + 3 = 2 + 3 = 5
 
b = x + 10 = 2 + 10 = 12

c = x + 11 = 2 + 11 = 13

All 3 sides are positive, so x = 2 is the answer.

Edwin