SOLUTION: Stumped once again!!! 2x^2-5x-2 =0 How is the quadratic formula used to solve this equation? Anyone know?

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Question 328580: Stumped once again!!!
2x^2-5x-2 =0
How is the quadratic formula used to solve this equation? Anyone know?
Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


For





For your problem:

, , and

so



Just do the arithmetic and simplify.

John

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic formula for finding roots is
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
You need to know what a, b, and c refer to
This formula assumes that the equation is in the form
ax%5E2+%2B+bx+%2B+c+=+0
Your equation is already in this form:
2x%5E2-5x-2+=0
a+=+2
b+=+-5
c+=+-2
plugging in the numbers:
x+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A2%2A%28-2%29+%29%29%2F%282%2A2%29+
x+=+%285+%2B-+sqrt%28+25+%2B+16+%29%29%2F4+
x+=+%285+%2B-+sqrt%2841%29%29%2F4
I'm guessing that c should be 2 and not -2. That would make the
solution
x+=+%285+%2B-+sqrt%2825+-+16%29%29%2F4
x+=+%285+%2B-+sqrt%289%29%29%2F4
x+=+%285+%2B+3%29%2F4
x+=+2
and
x+=+%285+-+3%29%2F4
x+=+1%2F2
These are the 2 roots.
If c+=+-2 for real, then stick with x+=+%285+%2B-+sqrt%2841%29%29%2F4