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that was 2x^2

\(\large\color{black}{ \displaystyle 2x^2 -7x + 3 = 0 }\)

what can i multiply to get 3 that i can add to get 7...nothing right?

the discriminant is a positive integer and that means that you can factor your equation.

so i use the formula \[-b \pm \sqrt{b-4(a)(c)/2}\]

-7/4 +or- 25/4?

hmmm

i don't think it is correct

let me redo

also, wouldn't factoring be a more efficient method than quadratic formula ?

but, if you want to use the quadratic formula, then sure, go ahead...

i tried factoring, and what i am being taught, is teaching the quadraic formula

ok, what I will suggest is to do it both methods.

show me how to factor it

ok.

Have you ever factored an equation before?

yes

how did you do it.
was it like you were sort of "guessing" ?

if so, then we can start doing the same thing.... it would be hard

wait no i didnt

should i divide all by 2?

im trying to do....\[2x ^{2} -?\pm?+ 3....\]

onc ei get that then i can do (2x )(x )

for example
\[x ^{2} + 4x +4.... would then be x^2 +2x +2x + 4\]

yes, I did that above

but -4x, times 3x doesnt equal 3?

i rewrote -7x as -4x-3x