Find two consecutive even integers such that the square of the smaller is 10 more than the larger.

Let the two consecutive even integers be x and x+2.

Then x^2 = (x+2) + 10

Simplifying, x^2-x -2 - 10 = 0, or x^2 - x - 12 = 0.  Factoring,

(x-4)(x+3) = 0.  Solving for x:  x=4 and x= -3.

If the first number is x = 4, then the second is x+2=4+2=6.

But wait!  If the first number is -3, then the second is x+2 = -3+2 = -1.

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