If the product of two consecutive even integers is 76 less than 10 times their sum, how do you find the two integers?

Apr 25, 2016

There are two possible solutions - $4$ and $6$ or $14$ and $16$

Explanation:

First put the statement into mathematical terms. Let $x$ be the first integer, then the second one is $x + 2$.

Then the product $x \left(x + 2\right) = 10 \left(x + \left(x + 2\right)\right) - 76$

Expanding,
${x}^{2} + 2 x = 20 x + 20 - 76$
${x}^{2} - 18 x + 56 = 0$
$\left(x - 14\right) \left(x - 4\right) = 0$

Then $x = 14$ or $x = 4$ and the second integer is $16$ or $6$