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.

Do NOT accept these answers until after you've checked them!