Exercise PhysicsWiki: Difference between revisions
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<math> x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} </math> (Sandy Simmons) | <math> x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} </math> (Sandy Simmons) | ||
<math>e^{i\pi} + 1 = 0</math> (Steve Honeywell) |
Revision as of 01:10, 30 August 2009
$$a=b$$
(posted by TerriC, Group 2)
(Zach McDargh)
(posted by KimberlyWynne 19:05, 29 August 2009 (EDT))
(Sandy Simmons)
(Steve Honeywell)