A community for students.
Here's the question you clicked on:
 0 viewing
ladiesman217
 3 years ago
medals and fans rewarded
ladiesman217
 3 years ago
medals and fans rewarded

This Question is Closed

ladiesman217
 3 years ago
Best ResponseYou've already chosen the best response.0@ajprincess @nubeer @.Sam. can you help me please

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.2\(3x^2=33x+24\) \(3x^233x+24=0\) Use quadratic formula. \[x=\frac{ b\pm\sqrt{b^24ac} }{ 2a }\] the values of a, b and c can be found by comparing the equation \(3x^233x+24=0\) with the general equation \(ax^2+bx+c=0\). Does that help? @ladiesman217

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0**24 dividing the entire equation by 3 would simplify calculations

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.2it is actually \(3x^233x24=0\). am sorry abt the mistake

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.2ya. first factor out 3 and then divide by 3 both sides. After that use quadratic formula:)

ladiesman217
 3 years ago
Best ResponseYou've already chosen the best response.0the equation i got was\[\frac{ (3)\sqrt{3^{2}4(3)(24)} }{ 2*3 }\]

ladiesman217
 3 years ago
Best ResponseYou've already chosen the best response.0this is the correct equation right \[\frac{ (33)\sqrt{33^{2}4(3)(24)} }{ 2*3 }\]

ajprincess
 3 years ago
Best ResponseYou've already chosen the best response.2it will be \[\frac{ (33)\sqrt{(33)^24*3*(24)} }{ 2*3}\]

ladiesman217
 3 years ago
Best ResponseYou've already chosen the best response.0i got\[\frac{ 11\pm \sqrt{153} }{ 2 }\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.