- #1

kimjoc

- 1

- 0

x= (-b± [itex]\sqrt{b^2-4ac}[/itex])/2a

Am I missing something? I mean the value underneath the square root turns out to be a negative number, so technically this equation would not have worked right? It's from a textbook by the way. This is actually an analytical chemistry application question.

x= -6.8*10

^{-4}± [(6.8*10

^{-4})

^{2}- (4)(6.8*10

^{-4})]

^{0.5}/(2)(1)