quadratic formula

i'm trying to use the quadratic formula for $\displaystyle 3x^2 - 12x + 5 = 0$

and got it to $\displaystyle x = \frac {12 \pm \sqrt {-12^2 - 60}}{6}$ not sure if i'm going about it right so would like someone to point me in the right direction if possible

thanks

Yes you are very correct.
But my teacher insisted on being picky so in that sentiment I'll just say that your notation, instead of http://www.mathhelpforum.com/math-he...dac4037a-1.gif

Should be

$\displaystyle x=\frac{12\pm{\sqrt{(-12)^2-60}}}{6}$

The addition of brackets removes confusion, that's what my teacher used to say.

ah thanks, the answer the book gives is $\displaystyle x = 2 \pm \sqrt {\frac{7}{3}}$ so i can't see how you'd arrive at that from the answer i was at, could you show me the way to get to it please? there are also meant to be rackets under the root sign around the 7/3 like so: (7/3) but i don't know how to write it out using the brackets like that

Quote:

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Originally Posted by mark

i'm trying to use the quadratic formula for $\displaystyle 3x^2 - 12x + 5 = 0$

and got it to $\displaystyle x = \frac {12 \pm \sqrt {-12^2 - 60}}{6}$ not sure if i'm going about it right so would like someone to point me in the right direction if possible

thanks

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$\displaystyle a = 3 $

$\displaystyle b = -12$

$\displaystyle c = 5$

$\displaystyle x = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)}$

$\displaystyle x = \frac{12 \pm \sqrt{144 - 60}}{6}$

$\displaystyle x = \frac{12 \pm \sqrt{84}}{6}$

$\displaystyle x = \frac{12 \pm \sqrt{4 \cdot 21}}{6}$

$\displaystyle x = \frac{12 \pm 2\sqrt{21}}{6}$

$\displaystyle x = \frac{6 \pm \sqrt{21}}{3} = 2 \pm \frac{\sqrt{21}}{3}$

note that $\displaystyle \sqrt{\frac{7}{3}} = \frac{\sqrt{7}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{21}}{3}$