SOLUTION: I am supposed to solve: x^2+14x-4=0 Is the answer x=0.28,14.28 I don't think that's right

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Question 235865: I am supposed to solve:
x^2+14x-4=0
Is the answer x=0.28,14.28
I don't think that's right
Found 2 solutions by checkley77, Edwin McCravy:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
YES.
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x^2+14x-4=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x=(-14+-sqrt[14^2-4*1*-4])/281
x=(-14+-sqrt[196+16])/2
x=(-14+-sqrt212)/2
x=(-14+-14.56)/2
x=(-14+14.56)/2
x=.56/2
x=.28 ans.
x=(-14-14.56)/2
x=(-28.56)/2
x=-14.28 ans.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B14x-4=0
Is the answer x=0.28,14.28
I don't think that's right 


Checkley said yes, so let's find out for sure:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ 

with a=1, b=14, c=-4

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ 

 

x+=+%28-14+%2B-+sqrt%28196%2B16+%29%29%2F2+

x+=+%28-14+%2B-+sqrt%28212+%29%29%2F2+

x+=+%28-14+%2B-+sqrt%284%2A53+%29%29%2F2+ 

x+=+%28-14+%2B-+2%2Asqrt%2853+%29%29%2F2+

x+=+%28-14%29%2F2+%2B-+%282%2Asqrt%2853+%29%29%2F2+

x+=+-7+%2B-+sqrt%2853+%29+

Using the +,

x+=+-7+%2B+sqrt%2853+%29+

Using a calculator:

x+=+-7+%2B+7.280109889+=+0.280109889

That rounds off to 0.28, so you're right on that one.


Using the -,

x+=+-7+-+sqrt%2853+%29+

Using a calculator:

x+=+-7+-+7.280109889+=+-14.280109889

That rounds off to -14.28.  Uh oh! You missed the
sign on that one. Checkley didn't notice.
But you were half right!

Edwin