SOLUTION: Find the roots of 4n2+7n-3=0 correct to two decimal places

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Question 1107194: Find the roots of 4n2+7n-3=0 correct to two decimal places
Found 3 solutions by Fombitz, Boreal, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use the quadratic equation,
n+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
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a=4
b=7
c=-3

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
4n2+7n-3=0
n=(1/8)(-7 +/- sqrt(49+48)); sqrt 97=9.85
n=2.85/8=0.36
n=-16.85/8=-2.11
graph%28300%2C300%2C-5%2C5%2C-5%2C5%2C4x%5E2%2B7x-3%29

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the roots [of 4n^2+7n-3=0 correct to two decimal places
Use the binomial formula to get:
n = [-7+-sqrt(7^2-4*4*-3)]/(2*4)
n = [-8 +- sqrt(30)]/8
n = [-1+ sqrt(30)/8] or n = [-1-sqrt(30)/8]
n = -0.32 or n = -1.68
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Cheers,
Stan H.
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