Help meee /.<
Solve this quadratic equation using the quadratic formula. \[3x^2 + 5x + 1 = 0\]
oh i literally just did this
Good for you >.>
I will leave it to the professionals, since I am a mere amateur at this
You have an equation in the form \[\large ax^2 + bx + c=0\] The quadratic formula states \[\large x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Yeah a=3 b=5 c=1 aaaa what' snext, it's stuck in my brain
Exactly! so now you just plug in each of those numbers for their corresponding variables a = 3 b = 5 c = 1 \[\large \frac{-(5) \pm \sqrt{(5)^2 - 4(3)(1)}}{2(3)} \]
Oh thanks, let me do this :3
OKay
Okay :D
DAMN IT DIDN'T POST! >.<
Lol uh oh...start over! >:D lol
sUREEE
Okay \[x=\frac{ 5-\sqrt{13} }{ 6 }\]
Yay it works xP
hmm, it does? Because that should be a -5 on top there :)
wait what o.o
o.m.g. well okay let me draw it
Well remember the equation was \(\large x = \frac{\color \red{-}5 \pm \sqrt{(5)^2-4(3)(1)}}{2(3)}\)
Yeah it was :3
So when you simplify that down you get \[\large x = \frac{-5 \pm \sqrt{13}}{6}\] WHICH OF COURSE!!!! you then split up into 2 equations since you have a \(\large \pm\) sign So your 2 answers are then?
YAASS THAT'S WHAT I GOT
Oh lol okay so basically i get\[\frac{ 5-\sqrt{13} }{ 6 }, \frac{ 5+\sqrt{13} }{ 6 }\]
Argh...where is that negative sign in front of the 5 missy! >.<
UGHHH okay iyt's stays there imaginary :3
Lol okay I'll deal with that XD as long as it's there in some sense haha
Fine fine fine take it\[\frac{ -5\sqrt{13} }{ 6 },\frac{ 5+\sqrt{13} }{ 6}\]
Lol I'm just going to type it for you XD :P \[\large \frac{-5 + \sqrt{13}}{6} , \frac{-5 - \sqrt{13}}{6}\]
Oh damn I made a mistake in the second one again x'P Sorry LOL thansk
Lol I just wanted to close the case XD \( \LaTeX \) can be terrible sometimes haha
Oh yea yeaa, liek i was trying to post it, and turnsw out, it justposted "okay" >.>
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