Question

solve using quadratic formula
3x^2+12x+8=0

Answer 1

a=3 b=12 c=8
-12±{12^2-4(3)(8)}/2(3)
-12±{144-96}/2(3)
-12±{48}/6

Answer 2

i used{} to represent square root symbol. im not sure if i did it right cos i cannot finihs it

Answer 3

whats the final answer?

Answer 4

im not sure because their is no square root of 48?

Answer 5

\[(-12+4\sqrt{3})/6\]
and
\[(-12-4\sqrt{3})/6\]

Answer 6

4 square roots of thre is equal to 48?

Answer 7

how do i solver for those next two equations then?

Answer 8

finally you will get
\[(-6+2\sqrt{3})/3\]
and
\[(-6-2\sqrt{3})/3\]

Answer 9

yup\[4\sqrt{3}=\sqrt{48}\]

Answer 10

(−6+2√3)/3=(−2+2√3)/1
(−6−2√3)/3=(−2−2√3)/1

Answer 11

nope

Answer 12

well hwat should i have done

Answer 13

Sorry I was at lunch. Let me take a look

Answer 14

Ok starting from what you had here:
\[\frac{-12 \pm \sqrt{48}}{6}\]\[=-\frac{12}{6} \pm \frac{4\sqrt{3}}{6}\]\[=-2 \pm \frac{2\sqrt{3}}{3}\]

Answer 15

ok

Answer 16

So I find it is often easier to break the fraction up into two then simplify each part

Answer 17

ok

Answer 18

hello?:(

Answer 19

hello what? Did you have another question?

Answer 20

how do i finish the problem?

Answer 21

polpak's answer is better than mine

Answer 22

whats the answer? im so confused

Answer 23

both are correct

Answer 24

those arent the answers thoguh usually its x= # or #

Answer 25

One is the + version. One is the - version.

Answer 26

\[-2+(2\sqrt{3}/3)\]
and
\[-2-(2\sqrt{3}/3)\]

Answer 27

\[x = -2 + \frac{2\sqrt{3}}{3}\]or\[x = -2 - \frac{2\sqrt{3}}{3}\]

That's what\[x=-2\pm \frac{2\sqrt{3}}{3}\]
means

Answer 28

i entered x= -2+2√3/3 or -2-2√3/3 and it was incorrect

Answer 29

Dunno what to tell you, those are the right answers. Could also look like:
\[\frac{2}{3}(-3 +\sqrt{3})\]
and
\[\frac{2}{3}(-3 -\sqrt{3})\]

Answer 30

hmm ok well thanks anyway i guess