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Solve x2 + 6x + 7 = 0. Options are below!!!

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know quadratic formula ?
Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
\[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
How do I apply that to this problem? Like i'm having trouble setting it up
none of the options are correct b^2-4ac = 36-28 = 8 so there should be sqrt 8 in options or sqrt 2
couldn't u find, a,b,c ?
Those are the only options.......
then your question must be incorrect
check it
also see your previous question
for this its, (-6+- sqrt 8)/2 = (-3+- sqrt2)
Solve x2 + 6x + 7 = 0
for that (-3+- sqrt2)
Oh my goodnes!!!!1 I'm SO sorry
i see what occurred
the question is: Solve 3x2 + 4x = 2
right, 3x^2+4x-2=0 now apply the formula can u find,a,b,c first ?
3, 2, 4?
a = 3 b = 4 c = -2
yes. now what is b^2-4ac = ?
one moment
16 - 4(3)(-2)???
four times a c? or four times a times c?
yes, evaluate that
It equals 40
yes, now put all values in formula and tell me what u get for x.
This is how I set it up: -4+sq root 4^2-4(3)(-2)/2(3)
= 36
x1=[-4 + sqrt (40)] / (2*3) = ? x2=[-4 - sqrt (40)] / (2*3) = ?
got that ^ ?
Yes!! :)
got final answer then ?
-13.4 for the second
from the options ? 3.7 is not in options keep sqrt 40 as 2sqrt 10 only
option b
absolutely correct! got it ?
YES!!!!!!!!!!!! It makes sense! why couldn't I see that on my own?? Thank you!!!!!!

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