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

the two positive consecutive integers are 4 and 6.

Step-by-step explanation:

Let the smaller integer be s; then s^2 = (s + 2) + 10.

Simplifying, s^2 - s - 2 - 10 = 0, or

s^2 - s - 12 = 0.

Solve this by factoring:  (s - 4)(s + 3) = 0.

Then s = 4 and s = -3.

If the first even integer is 4, the next is 6.  We omit s = -3 because it's not even.

The smaller integer is 4.  Does this satisfy the equation s^2 = (s + 2) + 10?

4^2 = (4 + 2) + 10  True or False?

16 = 6 + 10 = 16.

True.

So the two positive consecutive integers are 4 and 6.