Question

Use the quadratic formula to solve the equation below.
-6x= -4x^2+18

A.X= 3 | x= -15
B.X= 1.5 | C=-3
C.X= 3|X=-3
D.X= -3 | X= -1.5

Answer 1

\[x= \frac{- ( 6 )^2 \pm \sqrt{ (-6)^2 - ( 4 )( 4 ) ( -18)} }{ 2 ( 4 )} \]

Answer 2

Can you please explain this to me? @Luis_Rivera

Answer 3

\[x= \frac{ 36\pm \sqrt{ -92 } }{8} \]

Answer 4

now find x

Answer 5

I don't understand how to do this. I got 2.6 the way I did it. Please explain

Answer 6

@Luis_Rivera

Answer 7

first, try to bring this -6x= -4x^2+18 in the form
ax^2+bx+c=0
can you ?

Answer 8

So I would move the equal sign to the end and leave to -6x^2 where it originally was but add the equal sign?

Answer 9

i am not sure i understand that fully, but tell me what equation you got ?? i'll verify it.

Answer 10

-6x^2+-4x^2+18=0

Answer 11

Its probably wrong

Answer 12

firstly, its 6x and not 6x^2 , right ?
also, if you want to move -6x to right,
you add 6x on both sides, so that on left you get
-6x+6x =0
got this ?

Answer 13

Oh yeah sorry -6x

Answer 14

Yeah

Answer 15

now will you please try again and tell me what equation you get ?

Answer 16

-4x^2+18=0
Because we took out the -6x
But would I add 6x to the 18?

Answer 17

you would add 6x on right side too, right ?
so
-4x^2+6x+18 = 0

Answer 18

Okay. So then would we us pemdas or a different step?

Answer 19

Compare your quadratic equation with \(ax^2+bx+c=0\)
find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\]
then the two roots of x are:
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 20

tell me what you get for a,b,c, first

Answer 21

I don't know how to find them. I never learned this.

Answer 22

How would I find them?

Answer 23

now ?