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- Jan 26, 2012

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I think you mean you gotSolve for the roots of 3x^2+12x +8=0. I got +-12 /48 over 6. (/ was the square root sign) yet the answer is the same but -84 instead of 48. What is it?

**-48**instead of 48, but either way let's go through solving this.

\(\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Looking at $3x^2+12x+8$ we see that $a=3$, $b=12$ and $c=8$. Plugging that into the above quadratic equation yields.

\(\displaystyle x = \frac{-12 \pm \sqrt{144-4(3)(8)}}{6}\)

Note that \(\displaystyle \sqrt{144-4(3)(8)}=\sqrt{144-96}=\sqrt{48}\) so the above line simplifies to the following.

\(\displaystyle x = \frac{-12 \pm \sqrt{48}}{6}\)

\(\displaystyle x = \frac{-12 \pm 4 \sqrt{3}}{6}=-2 \pm \frac{2 \sqrt{3}}{3}\)

Now to answer your original question, the 48 under the square root sign should be

**positive**. You are most likely only dealing with

*real numbers*so you can't take the square root of a negative is a good rule to remember.

If you have any other questions about this I invite you to register and ask here.