# How do you determine three consecutive even integers such that the first times the third, is 4 less than 12 times the second?

Apr 21, 2016

-2,0,2
or 10,12,14

#### Explanation:

First of all, lets call the integers $\left(x - 2\right) , \left(x\right) , \left(x + 2\right)$. We can do this because consecutive integers differ by 2. Now from the information we have, we can make an equation:
$1 s t \cdot 3 r d = 12 \cdot 2 n d - 4$
$\left(x - 2\right) \left(x + 2\right) = 12 \cdot \left(x\right) - 4$
${x}^{2} - 2 x + 2 x - 4 = 12 x - 4$
${x}^{2} - 4 = 12 x - 4$
${x}^{2} = 12 x$
${x}^{2} - 12 x = 0$
$x \left(x - 12\right) = 0$
Now you see that there are two solutions to this, when $x = 0$ and $x = 12$.
So our integers can be:
-2,0,2
or 10,12,14