anonymous 4 years ago Can you help me see where I am stuck finding the zeros of this quadratic equation?

1. anonymous

$3x^2 - 12x + 11 = 0$

2. anonymous

quadratic formula for this one use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=3,b=-12,c=11$

3. anonymous

My student solutions manual says I should obtain $(6 \pm \sqrt{3}) \div 3$

4. anonymous

indeed you should

5. anonymous

Let's walk through this, so I can see where I am tripping over my shoelaces.

6. anonymous

first you get $\frac{-12\pm\sqrt{(-12)^2-4\times 3\times 11}}{6}$

7. anonymous

Wait a sec. I see that I made an error in the first term under the radical. I will go back to the drawing board and re-attempt this one on my own.

8. anonymous

By the way, how are you producing the nice looking divisor line?

9. anonymous

\frac{a}{b} looks like $\frac{a}{b}$

10. anonymous

Got it. I made simple mistakes in my substitutions, then got hung up on trying to do algebra with unduly complicated figures.

11. anonymous

THANKS @satellite73, you got me unstuck!