Use a quadratic equation to find two consecutive even integers if their product is $168$

Call the odd integer between the two even integers $n$. The even integers are then $n-1$ and $n+1$, so that $$168=(n-1)(n+1)=n^2-1$$ so that $n^2=169$ etc.

Intuitively, if $x$ and $y$ are close to each other, their product should be close to the square of their average. If you distort a square by shortening one side while enlarging the other, the area wouldn't change much:

$x$ and $y$ are consecutive even integers so their average is the odd number inbetween.

$\sqrt{168} \approx 12.961$

Which is close to $13$, an odd number. Now all you have to do is check if $12 * 14$ is the solution.

An even integer $x$ is of the form $2n$, $n \in \mathbb{Z}$,and the next even integer is 2 more, so $y = 2n+2$.

So $2n(2n+2) = 168$ or $4n^2 + 4n - 168 = 0$ etc. Having $n$ we find $x$ and $y$.