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OpenStudy (anonymous):
step by step plzz
OpenStudy (anonymous):
3x^2+4x+3=0
3x^2+4x=0-3
3x^2+4x=-3
OpenStudy (saifoo.khan):
can't be.
OpenStudy (saifoo.khan):
simply use quadratic formula.
OpenStudy (anonymous):
Simply is a wingspan word for someone who doesn't understand.
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OpenStudy (eyust707):
do you know what the quadratic is?
OpenStudy (anonymous):
it is 2
OpenStudy (anonymous):
\[{-b \pm \sqrt{b ^{2}-4ac}}/{2a}\]
OpenStudy (eyust707):
yeah!
OpenStudy (anonymous):
it algebra
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OpenStudy (saifoo.khan):
\[\frac{-b \pm \sqrt{b ^{2}-4ac}}{2a}\]
OpenStudy (eyust707):
ok and the function is in the form..
ax^2 + bx +c =0
OpenStudy (anonymous):
how'd you get the division line?
OpenStudy (anonymous):
yeep
OpenStudy (eyust707):
so what does a = ? b=? and c =?
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OpenStudy (eyust707):
*he probably used parenthesis* i knew what you meant tho
hero (hero):
No real solutions
OpenStudy (eyust707):
hero, how does that help math dude?
OpenStudy (zarkon):
\[\text{if you do a google search for }\LaTeX\text{ you will find all the commands you desire.}\]
hero (hero):
It helps
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OpenStudy (anonymous):
3x^2+4x+3=0
Solve the quadratic equation by completing the square:
Divide both sides by 3:
x^2+(4 x)/3+1 = 0
Subtract 1 from both sides:
x^2+(4 x)/3 = -1
Add 4/9 to both sides:
x^2+(4 x)/3+4/9 = -5/9
Factor the left hand side:
(x+2/3)^2 = -5/9
Take the square root of both sides:
sqrt(x+2/3) = (i sqrt(5))/3
Eliminate the absolute value:
x+2/3 = -(i sqrt(5))/3 or x+2/3 = (i sqrt(5))/3
Subtract 2/3 from both sides:
x = -1/3 i (-2 i+sqrt(5)) or x+2/3 = (i sqrt(5))/3
Subtract 2/3 from both sides:
x = -1/3 i (-2 i+sqrt(5)) or x = 1/3 i (2 i+sqrt(5))
hero (hero):
Copied and pasted from somwhere
OpenStudy (eyust707):
haha y u gotta hate?
hero (hero):
It's not hate if it's the truth
hero (hero):
Truth doesn't hate
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OpenStudy (anonymous):
@hero : i am just helping him yaar!!!!!
hero (hero):
lol
OpenStudy (eyust707):
lol
hero (hero):
I'm lucky I'm not in the same room with her. My head would be gone by now.
OpenStudy (anonymous):
lol
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