SOLUTION: I have to solve this using the quadratic equation 3x^2+2x+5=0

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Question 243355: I have to solve this using the quadratic equation
3x^2+2x+5=0
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
3x^2+2x+5=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x=(-2+-sqrt[2^2-4*3*5])/2*3
x=(-2+-sqrt[4-60])/6
x=(-2+-sqrt-56)/6
x=(-2+-7.4833i)/6
x=-2/6+-7.4833i/6
x=-1/3+-1.2472i ans. [complex roots]
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YES I'M SURE IT'S NEGATIVE BECAUSE THE FIRST TERM IS -B WHICH IS -2 & 2*A OR 2*3=6. SO -2/6=-1/3.