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Mathematics 22 Online
OpenStudy (anonymous):

Solve: 0 = 6x^2 - 10x - 4.

OpenStudy (anonymous):

you could factor that equation.

OpenStudy (anonymous):

how

OpenStudy (anonymous):

http://www.purplemath.com/modules/factquad.htm

OpenStudy (anonymous):

x+1/3, x=2???

OpenStudy (anonymous):

x i meant to put x=

OpenStudy (anonymous):

rewrite that please

OpenStudy (anonymous):

x=-1/3, x=2????

OpenStudy (anonymous):

is that correct?

jimthompson5910 (jim_thompson5910):

You can factor, but let's use the quadratic equation to solve this \[\Large 0 = 6x^2 - 10x - 4\] \[\Large 6x^2 - 10x - 4 = 0\] \[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \[\Large x = \frac{-(-10)\pm\sqrt{(-10)^2-4(6)(-4)}}{2(6)}\] \[\Large x = \frac{10\pm\sqrt{100-(-96)}}{12}\] \[\Large x = \frac{10\pm\sqrt{196}}{12}\] \[\Large x = \frac{10+\sqrt{196}}{12} \ \text{or} \ x = \frac{10-\sqrt{196}}{12}\] \[\Large x = \frac{10+14}{12} \ \text{or} \ x = \frac{10-14}{12}\] \[\Large x = \frac{24}{12} \ \text{or} \ x = \frac{-4}{12}\] \[\Large x = 2 \ \text{or} \ x = -\frac{1}{3}\] So you are correct iloveboys, nice work

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

great answer man! provs

OpenStudy (anonymous):

i have anoion... i have a graph liker this..

OpenStudy (anonymous):

|dw:1337814199315:dw|the question is what quadratcic* equationis represtenesed below... than the graph

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