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

The product of these two integers is given as 224. Therefore, the equation becomes:

$x(x + 2) = 224$

Expanding the equation:

$x^2 + 2x = 224$

Rearranging it into standard quadratic form:

$x^2 + 2x - 224 = 0$

Now, we solve this quadratic equation using the quadratic formula. The general quadratic formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 2$, and $c = -224$. Plugging these values into the formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-224)}}{2(1)}$ $x = \frac{-2 \pm \sqrt{4 + 896}}{2}$ $x = \frac{-2 \pm \sqrt{900}}{2}$ $x = \frac{-2 \pm 30}{2}$

Now, solving for both cases:

1. $x = \frac{-2 + 30}{2} = \frac{28}{2} = 14$
2. $x = \frac{-2 - 30}{2} = \frac{-32}{2} = -16$

So, the two possible pairs of consecutive even integers are:

1. $x = 14$ and $x + 2 = 16$
2. $x = -16$ and $x + 2 = -14$

Thus, the two pairs of integers are $14$ and $16$, or $-16$ and $-14$.

Would you like more details or have any questions?

Here are some related questions to expand on this concept:

1. What is the process of factoring quadratic equations?
2. Can we solve quadratic equations without using the quadratic formula?
3. What is the significance of discriminants in quadratic equations?
4. How can we find the sum of consecutive even integers?
5. How can we generalize the formula for the product of consecutive odd integers?

Tip: Always check if a quadratic equation can be factored before using the quadratic formula—it might save time!