The sum of the square of two consecutive even integers is 244 find the integer and form quadratic equation

Let the first even integer be $x$.
Since the integers are consecutive even integers, the next even integer is $x+2$.
The sum of the squares of these two integers is 244.

$$x^2 + (x+2)^2 = 244$$

$$x^2 + (x^2 + 4x + 4) = 244$$ $$2x^2 + 4x + 4 = 244$$

Subtract 244 from both sides:

$$2x^2 + 4x + 4 - 244 = 0$$ $$2x^2 + 4x - 240 = 0$$

Divide the entire equation by 2 to simplify:

$$x^2 + 2x - 120 = 0$$

The quadratic equation is $\boxed{x^2 + 2x - 120 = 0}$.

We can factor the quadratic equation $x^2 + 2x - 120 = 0$. We need two numbers that multiply to -120 and add to 2. These numbers are 12 and -10.

$$(x+12)(x-10) = 0$$

This gives two possible values for $x$:

$$x+12 = 0 \Rightarrow x = -12$$ $$x-10 = 0 \Rightarrow x = 10$$

Case 1: If $x = 10$
The first integer is 10.
The second integer is $x+2 = 10+2 = 12$.
Check: $10^2 + 12^2 = 100 + 144 = 244$.

Case 2: If $x = -12$
The first integer is $-12$.
The second integer is $x+2 = -12+2 = -10$.
Check: $(-12)^2 + (-10)^2 = 144 + 100 = 244$.

The integers are either $\boxed{10 \text{ and } 12}$ or $\boxed{-12 \text{ and } -10}$.