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i have a few of these this is just the first that i need help with...

ik the forumla is this:
x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

well we should use complete the square method
you can move the constant term to the right side

x is the term correct?

\[ \huge\rm 4x^2 − 12x + 3 = 0\]
subtract 3 both sides

not just x
term are divided by plus or negative sign

uh... subtract the 4x^2 by 3 and the 12 by 3?

|dw:1443453772906:dw|
you can't combine -3 and 4x^2 they arn't like terms

im not sure what you do next after you get -3..

divide?

|dw:1443453908051:dw|
we should complete the square of the left part

\[\huge\rm \color{Red}{4x^2+12x}=-3\]
what is the common factor at left side ?

the x?

no just take out the common number not the varaible

uh ... im confused. am i supposed to take out the 4 and 12 and divide them?

besides the x i see nothing in common...

well it's 4x^2+12x
so we just need to take out the common number

what are the factors of 4 and 12 ?

1, 2, 4 ... 4 is best factor... then i divide both by four to get 3 right?

greatest number **

uh i said 4...

is there one greater than thta?

that*

i don't understand what u mean "divide `both` to get 3"

\[\huge\rm 4(x^2+3x)=-3\]when you take out 4 from 4 u will get 1

you still have the 12x

i was saying didvide the 4x^2 and 12x by the common factor to simplify the solution.

yes right.
i read it wrong. so sorry

\[\huge\rm 4(x^2+3x)=-3\]
take half of the x term and then square it
add square to the both sides

you have to x's which one do i take half of?

two*

first one is x^2
2nd one is x

so take half of the second one?

yes right

uh... ok so now you'll have- x+ 3?

thats not a negative sign

x+3 how ?
we should take half of 3
in other words just divide `b by 2`

how am i dividing 3x by 2?

/i thought you wanted me to take away x on both sides..

well first we need to divide b by 2
take square of (b/2)^2
then add (b/2)^2 to both sides

what is b in ur equation ?

the 3x?

just b

3 =b
not 3x x is variable

do we get rid of negative 3 with the positive 3

positive 3 ?? it's 3x you can't combine 3x with -3 they are not like terms

then your left with whats in blue right?

you need to convert left side in squared form

so its a bit of a leap here but i believe the simplified form would be -3 +- 2\[\sqrt{6}\] over 2?

everything over 2*

and how did you get that ?

well you told me that like terms cant combine

leaving negative 3 alone.

then you said something about b being divided by2? which i believe gave me that..

how did you write \[(x^2+3x)\] in squared form ?

3x^2? idk

ik your left with one x right?

thats where im lost though...

yea that's obvious we have to solve for x

then how you got the final x value ?

i got what?

here is another example \[(x^2+6x)=(x+\frac{ 6 }{ 2})^2=(x+3)^2\]

wait but isn't that similar to what i had?

similar to what ?

solve this to get the answer \[4(x+\frac{ 3 }{ 2})^2=-3+(\frac{ 3 }{ 2 })^2 \times 4\]

feel free to ask question about that^let me know if you have question on that

ok thank you

i had another one mind helping?

did you get this one ?

i believe i did

Solve x2 − 5x + 1 = 0 using the quadratic formula.

alright i'll try

i know the first step to this ...

can u make a new post bec this one is laggy

i have noticed

ye it keeps moving
hard to use latex on this post