Don't pay any attention to this, test for my own sake, and for my cousins help...

*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!**