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Mathematics 18 Online
OpenStudy (anonymous):

Solve. 3x2 - 7x - 2 = 0 is the answer -7 _+_ sq root 25 over 6?

jimthompson5910 (jim_thompson5910):

the discriminant should be d = b^2 - 4ac d = (-7)^2 - 4(3)(-2) d = 49 + 24 d = 73

OpenStudy (anonymous):

how would it be a positive? slightly confused

jimthompson5910 (jim_thompson5910):

(-7)^2 = 49 - 4(3)(-2) = -12*(-2) = 24

jimthompson5910 (jim_thompson5910):

so d = (-7)^2 - 4(3)(-2) turns into d = 49 + 24

OpenStudy (anonymous):

oh, I see now, thank you @jim_thompson5910

jimthompson5910 (jim_thompson5910):

np, so this means that your final answers should be \[\Large \frac{7\pm\sqrt{73}}{6}\]

OpenStudy (anonymous):

Correct, thank you!

OpenStudy (anonymous):

Solve. 3x2 - 2x - 3 = 0

OpenStudy (anonymous):

x = sq root 80 over 3?

jimthompson5910 (jim_thompson5910):

no

OpenStudy (anonymous):

let me revise... i made an error obviously

jimthompson5910 (jim_thompson5910):

\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \[\Large x = \frac{-(-2)\pm\sqrt{(-2)^2-4(3)(-3)}}{2(3)}\] \[\Large x = \frac{2\pm\sqrt{4-(-36)}}{6}\] \[\Large x = \frac{2\pm\sqrt{4+36}}{6}\] \[\Large x = \frac{2\pm\sqrt{40}}{6}\] I'll let you finish

jimthompson5910 (jim_thompson5910):

keep in mind that \[\Large \sqrt{40}\] can be simplified

OpenStudy (anonymous):

I GOT IT!!!!!!!!

OpenStudy (anonymous):

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