The square of an integer is 12 more than the integer. Find the integer

x + 12 = x^2

x^2 - x - 12 = 0
(x + 3)(x - 4) = 0

x + 3 = 0
x = -3
This obviously doesn't work.

x - 4 = 0
x = -4
This doesn't work, but since both the positive and negative values of integers yield the same value when squared, let's try 4.

4^2 = 16
This is 12 more than the original integer 4.

from Pink Floyd's

(x + 3)(x - 4) = 0
x = -3 or x=4

check: is (-3)^3 12 more than -3 ? yes!
is (4^) 12 greater than 4 ? Yes!

so the numbers are either 4 or -3

By George, you're right.. just a little bit of math errors on my part.

To find the integer, let's assign a variable to it. Let's call the integer "x".

According to the given information, the square of the integer is 12 more than the integer. We can express this as an equation:

x² = x + 12

To solve this equation, we need to bring all the terms to one side to make it a quadratic equation:

x² - x - 12 = 0

Next, we can factorize the quadratic equation or use the quadratic formula to find the solutions. In this case, we can factorize the equation as follows:

(x - 4)(x + 3) = 0

Setting each factor equal to zero gives us two possible solutions:

x - 4 = 0 --> x = 4
x + 3 = 0 --> x = -3

Therefore, the two possible integers are 4 and -3.