-x^2-2x+24=0
-x^2-2x+24=0
someone please help me
U can think that a=-1, b=-2, c=24 ok
Now, we star u solve the problem
With that problem we will have x always gives 2 answers
i dont know how to work it out, i miss class for a week so i dont have any notes
1) X=( -b + square delta T) /2a, 2) x= -b - square delta T ) / 2 a , so u will get can answer. U find delta T by the way b ^2-4ac
U do understand ?
use quadratic equation. do you know how to use it?
no im highly confused, could u just break it down for me
Can you factor it?
you could and leave the negative sign on the outside
nooo i dont think you can.....
no i have to solve it
You could do this thing: -x² - 2x + 24 = 0 -(x² + 2x - 24) = 0 x² + 2x - 24 = 0 Now you can factor it easily. Is this clear? Factoring it is faster way to solve. Faster than using quadratic equation.
You know how to factor? No?
Ok let me right it down for u to easy understand
Ok let galacticwaves xx did it
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a}\] and you know that \[ax^2+bx+c=0\] from the 2nd equation you can tell by looking at your problem what a, b, and c is
Exactly what I said, good jobs men
i still need help
now what you have to do is plug in a, b and c into the 1st equation i put up above haha thank you as long as we speak the same language we succeed
@jkjk1 now can you do that and let me see what you get for x
i cant i really need help
hmmm ok i will first tell you that a=-1 b=-2 and c=24 you can see how i go this from the 2nd equation i posted up, correct?
could u give me the answer, i have to go to class at 6:30
???
then from plug in the values of a, b, and c onto the quadratic equation which is the 1st equation i posted up \[x=\frac{ -(-2) \pm \sqrt{(-2)^2-4(-1)(24)} }{ 2(-1) }=\frac{ 2 \pm \sqrt{4+96} }{ -2 }=\frac{ 2 \pm \sqrt{100} }{ -2 }\]
and you get x=-6 and x=4
You would be grateful if you know how to factor; in this way you don't have to do all these work lol...
He can u factor geeky, can u tell ur way?
let me see how you can factor this out because i don't think it is possibl
Join our real-time social learning platform and learn together with your friends!