Question

Am I right?

Question is in comments.

Answer 1

bring across the 8

Answer 2

you get 2x^2 #+9x-8=0

Answer 3

think @Secret-Ninja ninja can solve from there!

Answer 4

\[My~Work: \]
\[2x^2+9x=8 ~~~ turns~into~~~~~ 2x^2+9x-8=0\]

\[x= \frac{ 9\pm \sqrt{9^2-4(2)(8)} }{ 2(2) }\]

\[x=\frac{ 9\pm \sqrt{81-64} }{ 2(2) }\]

\[x=\frac{ 9\pm \sqrt{17} }{ 4 }\]

\[x=\frac{ 9+4.1231 }{ 4 } ~~~~~~~~~~~~~~~~~~ x=\frac{ 9-4.1231 }{ 4 }\]

\[x=\frac{ 13.1231 }{ 4 } ~~~~~~~~~~~~~~~~~~~~~~ x=\frac{ 4.8769 }{ 4 }\]

\[x = 3.280775~~~~~~~~~~~~~~~~~~~~~~~x=1.219225 \]

Answer 5

The closest answer to my work was Choice C.

Answer 6

@KamiBug

Answer 7

Sorry, those 9's were supposed to be negative, but you get the idea.

Answer 8

:p

Answer 9

I think @Secret-Ninja work is correct

Answer 10

pwerhaps there is a difference in round-off?

Answer 11

D: I got it wrong. :O :[

Answer 12

can you show the the answers you were given?

Answer 13

you put in -c instead of c

Answer 14

\[2x^2+9x-8=0 \\ x=\frac{-9 \pm \sqrt{9^2-4(2)(-8)}}{2(2)}\]
do you see what I mean?

Answer 15

\[x=\frac{-9 \pm \sqrt{81+64}}{4} \\ x=\frac{-9 \pm \sqrt{145}}{4}\]

Answer 16

Freckles seems to have this

Answer 17

*facepalm* I sometimes just feel so stupid >.<

Answer 18

Stop that @Secret-Ninja , the no.1 idea you have to keep in your head is "confidence"!

Answer 19

Ninjia work: \(\huge x= \frac{ 9\pm \sqrt{9^2-4(2)(8)} }{ 2(2) }\)
while freckles one: \(\huge x= \frac{ \color{red}{-}9\pm \sqrt{9^2-4(2)(8)} }{ 2(2) } \)
that makes waaaaaaaaaaaaay different results. hehehe

Answer 20

see there you go, they are arguing..

Answer 21

Yep. :/ Oh well, no going back now lol. Thanks for clearing it up though guys :]

Answer 22

difference is a minus

Answer 23

simple mistake!

Answer 24

well that isn't what I said...

Answer 25

\[x=\frac{-9 \pm \sqrt{9^2-4(2)(-8)}}{2(2)}\]

Answer 26

she changed the 9's to -9 she said
so that wasn't her problem
it was she entered in 8 instead of -8

Answer 27

and i didn't know anyone was arguing

Answer 28

is this finished?

Answer 29

I mean you have two very proficient members dealing with this problem

Answer 30

@freckles and @Loser66

Answer 31

?

Answer 32

This question is closed... Its all over. O.o