# Math Help - Quadratic formula question

Good morning everyone, I am having some trobule with the following equation.
This is what I have done so far.

4x^2+6x-2=0
I am using ax^2+bx+c=0
so
a=4
b=6
c=-2

$x=\frac{-(6)\pm\sqrt{6^2-4(4)(-2)}}{2(4)}$
$x=\frac{-(6)\pm\sqrt{36+32}}{8}$
$x=\frac{-(6)\pm\sqrt{68}}{8}$
$x=\frac{-(6)\pm\sqrt{8.25}}{8}$
$x=\frac{-(6)+8.25}{8},\frac{-(6)-8.25}{8}$
$\frac{-14.25}{8},\frac{2.25}{8}$

Does this look right to you?

2. Originally Posted by EricS
Good morning everyone, I am having some trobule with the following equation.
This is what I have done so far.

4x^2+6x-2=0
I am using ax^2+bx+c=0
so
a=4
b=6
c=-2

$x=\frac{-(6)\pm\sqrt{6^2-4(4)(-2)}}{2(4)}$
$x=\frac{-(6)\pm\sqrt{36+32}}{8}$
$x=\frac{-(6)\pm\sqrt{68}}{8}$
$x=\frac{-(6)\pm\sqrt{8.25}}{8}$
$x=\frac{-(6)+8.25}{8},\frac{-(6)-8.25}{8}$
$\frac{-14.25}{8},\frac{2.25}{8}$

Does this look right to you?
They look right.

Note that $\sqrt{68} \approx 8.25$ but if you have already rounded and converted into a decimal why not keep going?

$\frac{-14.25}{8} \approx -1.78,\frac{2.25}{8} \approx 0.28$

3. ## on another note

$
x=\frac{-(6)\pm\sqrt{68}}{8}
$

but that is just me.

4. Originally Posted by TheEmptySet

$
x=\frac{-(6)\pm\sqrt{68}}{8}
$

but that is just me.
Or even better:
$x = \frac{-3 \pm \sqrt{17}}{4}$

Whether you want an exact answer or decimals depends, of course, on why you are solving the equation.

-Dan

5. Awsome. I cant believe I got that one right! well sort of atleast, but i guess i'm on the right track.