# Three consecutive even integers are such that the square of the third is 76 more than the square of the second. How do you determine the three integers?

One can express the three consecuitve even numbers as $2 x , 2 x + 2 , \mathmr{and} 2 x + 4$. You are given that ${\left(2 x + 4\right)}^{2} = {\left(2 x + 2\right)}^{2} + 76$. Expanding the squared terms yields $4 {x}^{2} + 16 x + 16 = 4 {x}^{2} + 8 x + 4 + 76$.
Subtracting $4 {x}^{2} + 8 x + 16$ from both sides of the equation yields $8 x = 64$. So, $x = 8$. Substituting 8 for x in $2 x , 2 x + 2 , \mathmr{and} 2 x + 4$, gives 16,18, and 20.