Question

solve by completing the square
0=6x2+5x-5

Answer 1

5=6x^2+5x
5=x(6x+5)
how can I go about from here?

Answer 2

instead of rearranging you could have solved for x by using the quadratic equation
do you know it?

Answer 3

6x^2+5x-5=0?

Answer 4

sorry I tried but there was nothing to factor for...
usually this becomes something like
(x+/- something)(x+/- something)=0

Answer 5

and if its the one with the weird suare rooting 4ab something I didn't learn that, and I would appreciate it if yu can show me the formula. I was sick the day the teacher taught it, and she told me to "ask my friends" or "look it up"

Answer 6

yeah I meant the weird one which you didn't learn
so we can go through it together

Answer 7

I would greatly appreciate that... having a test after the break and I'm kind of screwed

Answer 8

ok the formula is\[\frac{ -b pm sqrt{b^{2}-4ac} }{ 2a}\]

Answer 9

well given the fact you are asked to use completing the square that's what you should do

divide every term by 6

\[x^2 + \frac{5}{6}x = \frac{5}{6}\]

find half the coefficient of x and then square it...

add it to both sides of the equation...

then you will have a perfect square on the left.

Answer 10

the system is playing up on my end but you can figure out the formula on line

Answer 11

so if the value of a is 6
and b is 5
and c is -5
you slot that into the formula

Answer 12

whoa what's the pm?
@campbell_st I wasn't told to comple te the square, but that was the only way I thought of going since I could not factor. I should go do the other thing I'm assuming

Answer 13

pm is plus/minus

Answer 14

oh... ok not it makes more sense..
so it's
(-1+/-sqrt(1-4(8)(-5))/16

Answer 15

sorry the problem was
0=6x^2+x-5

Answer 16

Here @nos since @Mariam.A since you were having trouble posting it:
if the equation is:
\[{ax^{2}+bx+c=0}\]\[x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\]

Answer 17

so it goes
-1+/-sqrt(-1-(4(6)(-5))/12
-1+/-sqrt(-1+120)/16
-1+/-sqrt(119)/16
and
-1+/-10sqrt(19)/16
-1/16+/-5sqrt(19)/8

Answer 18

more or less?

Answer 19

you messed up the b^2. your first line should be:
-1+/-sqrt(1-(4(6)(-5))/12

Answer 20

oh so the sqrt(121) is the correct answer? then it would be 11

Answer 21

-1+/-11/12
-12/12=-1
10/12=5/6

Answer 22

yes. The equation you gave is actually factorable:
\[0=6x^2+x-5\]\[0 = (6x-5)(x+1)\]

Answer 23

<I just need to try those answers>

Answer 24

wow. that lost the purpose it's ok at least it teached me how to do the weird thing, and I'm actually thankful for it.

Answer 25

I should check my answers, but thanks for the steps on the equation :)

Answer 26

yes that weird thingy that you said is actually very helpful so it would be wise to learn how to do it.