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Mathematics 22 Online
OpenStudy (neverendingcycle):

Solve the quadratic equation using the quadratic formula. 6x²- 9x + 2 = 0

RhondaSommer (rhondasommer):

I can help

OpenStudy (youngjuvie):

\[x=\frac{ 9\pm \sqrt{33} }{ 12 }\]

RhondaSommer (rhondasommer):

\[(-b +/- √b^2-4ac)/2a\]

OpenStudy (youngjuvie):

by Solve the equation for x by finding a,b, and c of the quadratic then applying the quadratic formula

OpenStudy (neverendingcycle):

is this a middle school question?

OpenStudy (youngjuvie):

decimal form x≈1.22871355,0.27128645

RhondaSommer (rhondasommer):

6x²- 9x + 2 = 0 a b c

RhondaSommer (rhondasommer):

not realy more like 8th and 9th grade depending how advanced you are

OpenStudy (neverendingcycle):

im in 8th grade and i have this question on a test but I never read about this in my studies.

RhondaSommer (rhondasommer):

@youngjuvie you are not explaing your answer. also; you answer isnt entirely correct. please stop or i will report you

OpenStudy (youngjuvie):

what grade ae u in

OpenStudy (neverendingcycle):

8th

RhondaSommer (rhondasommer):

oh cool yeah I can help you with this

RhondaSommer (rhondasommer):

so you know how a quadratic equation is written as this right? \[ax^2+bx+c\]

OpenStudy (neverendingcycle):

no i don't even know what quadratic means

RhondaSommer (rhondasommer):

involving the second and no higher power of an unknown quantity or variable

RhondaSommer (rhondasommer):

its a graph basically

OpenStudy (neverendingcycle):

i haven't learned this yet, but it is on a test

RhondaSommer (rhondasommer):

idk your teacher is wierd

OpenStudy (neverendingcycle):

okay

OpenStudy (neverendingcycle):

home schooled

OpenStudy (neverendingcycle):

my computer is my teacher

RhondaSommer (rhondasommer):

so the equations are this \[ax^2+bx+c\] with \[a,b,c\] as your co-efficients(numbers)

OpenStudy (neverendingcycle):

okay

RhondaSommer (rhondasommer):

lol me; but i started just this year

OpenStudy (neverendingcycle):

i started last year

RhondaSommer (rhondasommer):

\[6x²- 9x + 2 = 0\] is your equation with a=6 b=-9 and c=2

RhondaSommer (rhondasommer):

nice; lmk if you need help with anything ever ^_^ 8th was my best year with grades

OpenStudy (neverendingcycle):

okay i will fan you so i can remember your name

OpenStudy (neverendingcycle):

thanks for your help!

RhondaSommer (rhondasommer):

so you have this as your quadratic "equations" \[-b+√(b^2-4ac)/2a\]\[-b-√(b^2-4ac)/2a\]

OpenStudy (neverendingcycle):

okay

RhondaSommer (rhondasommer):

im helping you get the answer to this...did you not want that?

OpenStudy (neverendingcycle):

no i do

RhondaSommer (rhondasommer):

oh okay :)

OpenStudy (neverendingcycle):

:)

RhondaSommer (rhondasommer):

a=6 b=-9 and c=2 so plug those numbers into the first equation

OpenStudy (neverendingcycle):

okay

RhondaSommer (rhondasommer):

\[−b+√(b^2−4ac)/2a\] \[(9+√(-9)^2-4(6)(2))/2(6)\]

OpenStudy (neverendingcycle):

ohh okay that makes more since

RhondaSommer (rhondasommer):

so solve that while i solve the second equation :)

OpenStudy (neverendingcycle):

okay

RhondaSommer (rhondasommer):

OpenStudy (neverendingcycle):

i got 14 for my equation

RhondaSommer (rhondasommer):

RhondaSommer (rhondasommer):

you shouldnt have gotten 14 :/

OpenStudy (neverendingcycle):

well i stink at equations then :/

RhondaSommer (rhondasommer):

thats what you should have gotten... :/ i dont think you stink I think you needed an easier problem XD

OpenStudy (neverendingcycle):

yes i think that to

RhondaSommer (rhondasommer):

would you be wiling to do an easier problem that has a whole number as an equation?

RhondaSommer (rhondasommer):

not equation answer*

OpenStudy (neverendingcycle):

um, i think I will have to say not to that. Not because I don't like you, but because this is like a "what grade level is your math at" type of test so I don't think I will have to deal with that type of problem for a while. Thanks for all of your help though!

RhondaSommer (rhondasommer):

yeah no prob :) tag me if you need anything

OpenStudy (neverendingcycle):

okay, bye

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