Mirror due Midnight!!

[mustang]

Problem 32.
A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an ulluminated real object to be accurately traced.
If you have a concave mirror with a focal length of 16 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? In units of cm.
Note: I figured out that the formula I should use is 1/f=1/p+1/q
so would it be 1/f=2/-q+1/q
If so i need help to go on if not i need more help please this is due by Midnight.
f=16

[Severian596]

Problem 32.
A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an ulluminated real object to be accurately traced.
If you have a concave mirror with a focal length of 16 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? In units of cm.
Note: I figured out that the formula I should use is 1/f=1/p+1/q
so would it be 1/f=2/-q+1/q
If so i need help to go on if not i need more help please this is due by Midnight.
f=16
You didn't define q and p, but I think you're trying to say that p = 2q (image distance equals twice the object distance). Why put the 2 in the numerator, then? If p = 2q:

\frac{1}{f} = \frac{1}{2q} + \frac{1}{q}

\frac{2}{f} = \frac{1}{q} + \frac{2}{q}

\frac{2}{f} = \frac{3}{q}

q = \frac{(3)(f)}{2}= \frac{48}{2} = 24 cm