Vol. 201 No. 2 (2023): Dal Test Journal, Summer 2023

Articles

• This is a great article

  Gee Off

  • Article

• gsdfhgsdfh

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  • article

• Let's consider x = − b ± b 2 − 4 a c 2 a

  Gee Bee

  • PDF

• Something more complicated When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

  Gee Bee

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