Find 2 consecutive integers with a Product of 72

The of consecutive integers equals

We need two consecutive integers (n) and (n + 1) who have a product = 72

Setup relational equation:
We need to find two integers, n and n + 1 who have a product of 72
n * (n + 1) = 72
Multiplying through, we get n2 + n = 72
Rearranging the equation we get n2 + n - 72 = 0

Now that it is in Quadratic Format, determine a, b, and c:
a = 1, b = 1, and c = -72
Solution 1 = ½(-b + √b2 - 4ac)
Solution 1 = ½(-1 + √12 - 4 * 1 * -72)
Solution 1 = ½(-1 + √1 - -288)
Solution 1 = ½(-1 + √289)
Solution 1 = ½(-1 + 17)
Solution 1 = ½(16)
Solution 1 = 8

Determine Answers:
Solution 2 = Solution 1 + 1
Solution 2 = 8 + 1
Solution 2 = 9

Also, since the product of 2 negative #'s is positive, another solution is:
Solution 3 = (-1 * 8) * (-1 * 9)
Solution 3 = -1 * 8
Solution 3 = -8
Solution 4 = -1 * 9
Solution 4 = -9

Since 8 * 9 = -8 * -9 = 72, we have our solutions.