SOLUTION: can some one check this for me please? solve by using quadratic formula 5x^2+x+7=0 x=_+sqrt-2^2-5*5*7/2*5 (2)+sqrt(-4)5(7)/2(5) 2+16-49/2 2+sqrt33/2

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Question 80808: can some one check this for me please?
solve by using quadratic formula
5x^2+x+7=0
x=_+sqrt-2^2-5*5*7/2*5
(2)+sqrt(-4)5(7)/2(5)
2+16-49/2
2+sqrt33/2
Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(48546) About Me  (Show Source):
You can put this solution on YOUR website!
solve by using quadratic formula
5x^2+x+7=0
x=[-b+-sqrt(b^2-4ac)]/(2a)
x=[-1+-sqrt(1^2-4*5*7]/(2*5)
x=[-1+-sqrt(-139))]/(10)
x=(-1/10) + (1/10)sqrt(139)i or x=(-1/10) - (1/10)sqrt(139)i
============
Cheers,
Stan H.

Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
5x%5E2%2Bx%2B7+=+0 Apply the quadratic formula:x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x+=+%28-1%2B-sqrt%281%5E2-4%285%29%287%29%29%29%2F2%285%29 Simplify.
x+=+%28-1%2B-sqrt%281-140%29%29%2F10
x+=+%28-1%2B-sqrt%28-139%29%29%2F10
The solutions are the "complex" numbers:
x+=+%28%28-1+%2B+sqrt%28139%29i%29%29%2F10 and
x+=+%28%28-1+-+sqrt%28139%29i%29%29%2F10
If you need the approximate solutions:
x+=+-0.1%2B%281.1790%29i and
x+=+-0.1-%281.1790%29i to the nearest ten thousandth.