The product of two consecutive even integers is 48 find the integers

To start, define your variables, x will be the first even integer, and x+2 will be the greater of the two even integers (you would add 2 since even integers are 2 numbers apart). Then, since the product of the two integers is 48, you would set up an equation that looks like this: x(x+2)=48.
Once you simplify, you should get an equation that looks like a quadratic: 
x^2+2x=48. Now, to find the values of x, you have to solve for the roots or zeroes, which means you will have to set the equation to 0. To do this, subtract 48 from both sides and you get: x^2+2x-48=0.
Then, factor the equation and you get: (x+8)(x-6)=0
Now, to solve for x, set each factor to 0, which would look like this:
x+8=0
x-6=0
Now, solve each one for x and you get x=-8 and x=6
Then, to find the pairs for each integer, add 2 to each x and you get that the two sets of integers that equal 48 as their product are -8 and -6; and 6 and 8

Y=-2x-6 is the answer