SOLUTION: Can you please help me solve this? Please use the quadratic formula. 2x^2+x-5=0

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Question 34652: Can you please help me solve this? Please use the quadratic formula. 2x^2+x-5=0
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B1x%2B-5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A2%2A-5=41.

Discriminant d=41 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+41+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+41+%29%29%2F2%5C2+=+1.35078105935821
x%5B2%5D+=+%28-%281%29-sqrt%28+41+%29%29%2F2%5C2+=+-1.85078105935821

Quadratic expression 2x%5E2%2B1x%2B-5 can be factored:
2x%5E2%2B1x%2B-5+=+2%28x-1.35078105935821%29%2A%28x--1.85078105935821%29
Again, the answer is: 1.35078105935821, -1.85078105935821. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B1%2Ax%2B-5+%29

Hope this helps.
Cheers,
stan H.