# Prove that a converging lens has 2 object distances for which there exists an image

I've been struggling with this for a while now. I feel that proving this is actually way easier than what I think it is. I'm just going to copy paste the question here:

A light bulb (the object) is placed a distance of 753 mm from a screen. A convex lens with a focal length of 53 mm is placed between them. Show that there are two positions at which the lens can be placed so that an image is formed on the screen. Give the shorter object distance as your answer.

We have the equation 1/f = 1/p + 1/i . We did this in an experiment, the one image formed when the lens was midway between the screen and the object (light bulb) and the other formed when the lens was pretty far away. The lab-guy did say that you should get the equation quadratic, so that you have 2 answers. I just don't know what you can do to make this quadratic. I tried to sub in 1/f = 1/(p+d) + 1/i, then to say 1/(p+d) + 1/i = 1/p + 1/i, but this just gives d = 0 (in hindsight this should've been obvious). It ended with p+d=p, which could be written as p = p-d. I don't even know anymore.Also just a note, the light rays were not parallel when they entered the lens

HINT: what is the relationship between the object distance $$p$$ and the image distance $$i$$?
Ok I got it. First off, I totally misread the question and thought that the object distance is 753mm, when actually the 753mm is $$p$$+$$i$$. So yeah, $$p$$ = 753 - $$i$$. If you sub that into the equation 1/$$f$$ = 1/$$p$$ + 1/$$i$$, you get that $$i$$ = 695.63 mm or $$i$$ = 57.37 mm, which would make the lowest $$p$$ = 57.37 mm.