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)