SOLUTION: solve this please: x^2+5x-5=0. I'm stuck.

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Question 196901: solve this please: x^2+5x-5=0. I'm stuck.
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5x%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=%285%29%5E2-4%2A1%2A-5=45.

Discriminant d=45 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+45+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+45+%29%29%2F2%5C1+=+0.854101966249685
x%5B2%5D+=+%28-%285%29-sqrt%28+45+%29%29%2F2%5C1+=+-5.85410196624968

Quadratic expression 1x%5E2%2B5x%2B-5 can be factored:
1x%5E2%2B5x%2B-5+=+%28x-0.854101966249685%29%2A%28x--5.85410196624968%29
Again, the answer is: 0.854101966249685, -5.85410196624968. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B-5+%29