I saw this image as a joke, and wondered if the equation can be solved. Can anyone perhaps assist with a solution?
Disclaimer: The note is fictional. I do not have access to the card in the picture, and am not seeking to find a pin code for any card. I am merely interested if the equation can be solved. If someone can just confirm that it is indeed solveable and works out to a valid value without giving the actual answer, that would be enough for me.
Whoever wrote this is clearly not very good at math, since they put an $x^2$ twice, and an ambiguous "equation". I'm going to make an assumption that the problem is actually an integral
$$ I = \int \frac{x^3 - 3x^2 + 2x - 1}{x^2 - 4x + 4} dx $$
Performing long division on the integrand
$$ \frac{x^3 - 3x^2 + 2x - 1}{x^2 - 4x + 4} = x + 1 + \frac{2x-5}{x^2-4x+4} $$
Then partial fractions
$$ x + 1 + \frac{2x-5}{x^2-4x+4} = x + 1 + \frac{2(x-2) - 1}{(x-2)^2} = x + 1 + \frac{2}{x-2} - \frac{1}{(x-2)^2} $$ Integrating the above gives
$$ I = \frac{x^2}{2} + x + 2\ln|x-2| + \frac{1}{x-2} + C $$
Overall, it's pretty straightforward.