1. **State the problem:** Find two positive even consecutive integers such that the square of the smaller integer is 10 more than the larger integer. 2. **Define variables:** Let the smaller even integer be $x$. Since the integers are consecutive even numbers, the larger integer is $x+2$. 3. **Write the equation:** The square of the smaller integer is 10 more than the larger integer, so: $$x^2 = (x+2) + 10$$ 4. **Simplify the equation:** $$x^2 = x + 12$$ 5. **Rearrange to standard quadratic form:** $$x^2 - x - 12 = 0$$ 6. **Factor the quadratic:** $$x^2 - x - 12 = (x - 4)(x + 3) = 0$$ 7. **Solve for $x$:** $$x - 4 = 0 \Rightarrow x = 4$$ $$x + 3 = 0 \Rightarrow x = -3$$ 8. **Check for positive even integers:** $x = 4$ is positive and even, $x = -3$ is negative and odd, so discard $-3$. 9. **Find the larger integer:** $$x + 2 = 4 + 2 = 6$$ 10. **Verify the condition:** Square of smaller integer: $$4^2 = 16$$ Larger integer plus 10: $$6 + 10 = 16$$ Condition holds true. **Final answer:** The two positive even consecutive integers are $4$ and $6$.