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April 21, 2014

April 21, 2014

Posted by **Jon** on Saturday, March 1, 2008 at 7:29pm.

I don't have an answer but I made a table and plugged in these values for x respectfully: -1/4, -2/4, -3/4, -1, 0

Thats all I have

- Algebra II -
**No name**, Saturday, March 1, 2008 at 8:54pmsomehow it's hard to find the roots manually since

b^2 - 4ac <0

- Algebra II -
**Reiny**, Saturday, March 1, 2008 at 9:26pmfirst of all bring all terms to one side

2x^2 - 3x - 2 = 0

now let f(x) = 2x^2 - 3x - 2

did you notice that it factors?

f(x) = (2x+1)(x-2)

so the x-intercepts are -1/2 and 2

which then become the roots of the equation from above.

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