Question

A screen is placed a distance d = 43.0 cm to the right of a small...

A screen is placed a distance d = 43.0 cm to the right of a small object. At what two distances to the right of the object can a converging lens of focal length +5.10 cm be placed if the real image of the object is at the location of the screen? (Hint: Use the thin lens equation in the form s' = sf/s-f , so s'(s − f) = sf. Write s' = d − s and solve for s in the resulting quadratic equation.)

Find the shorter distance and the longer distance (these values are 5.913146565 and 37.08685343)

For this lens, what is the smallest value d can have and an image be formed on the screen?

The second half is the part I don't understand

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