trying to figure out how my teacher got this answer using the quadratic equation.

Problem: 3x^2 + 2x -6

This is how he did it in class

3x^2 + 2x - 6 = 0
x= -2 +- square root 4-4(3)(-6) / 2(3)

x= -2 +- square root 4 * 19 / 6

= -1/3 + square root 19/3

= square root 19-4 / 3

That is the final answer but I dont know how he got that.

2. $\displaystyle 3x^2 + 2x -6 = 0$

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

$\displaystyle x=\frac{-2 \pm \sqrt{4-4(-6)(3)}}{6}$
$\displaystyle x=\frac{-2 \pm \sqrt{76}}{6}$
$\displaystyle x=\frac{-2 \pm \sqrt{4*19}}{6}$
$\displaystyle x=\frac{-2 \pm 2\sqrt{19}}{6}$
$\displaystyle x=\frac{-1 \pm \sqrt{19}}{3}$

3. ok so using the same method

3x^2 + 2x -6.5

x= -2 +- square root 4-4(3)(-6.5) / 2 (3)

x = -2 +- square root 82 / 6

then I get stuck right here.

4. Originally Posted by winterjm
ok so using the same method

3x^2 + 2x -6.5

x= -2 +- square root 4-4(3)(-6.5) / 2 (3)

x = -2 +- square root 82 / 6

then I get stuck right here.
It would help, for starters, if you would use parenthesis. Your last line should read:
x = (-2 +- square root(82))/6

Right-o then. Since $\displaystyle \sqrt{82}$ can't be simplified you are done. There is nothing left to do for this one.

-Dan