Find 3 consecutive even integers such that the square of the largest number is 124 more than the square of the second largest number.

Let your 3 numbers equal X, X+2 and X+4 (because of the even integer boundary)

So, According to the problem, <largest number squared>$\displaystyle (x+4)^2 - 124 = (x+2)^2$<second number squared>

Expand:

$\displaystyle (x+4)(x+4) = (x^2+8x+16)-124 = x^2+4x+4$

Simplify both sides of the equations by removing the integers (+16 and -4 to the 124):

$\displaystyle
x^2+8x-112=x^2+4x
$

Take -112 to the right side to become positive:

$\displaystyle x^2+4x = x^2+112$

Minus $\displaystyle x^2$ and solve for x:

$\displaystyle 4x=112$
$\displaystyle
x=28
$
add original formulas to get the final 3 even integers of 28, 30, and 32!