SOLUTION: 2x^2+3x+5=125

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Question 22555: 2x^2+3x+5=125
Answer by Alwayscheerful(414) About Me  (Show Source):
You can put this solution on YOUR website!
You want it to be a quadratic equation, so it needs to be = to 0
2x%5E2%2B3x-120=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B3x%2B-120+=+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=%283%29%5E2-4%2A2%2A-120=969.

Discriminant d=969 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+969+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+969+%29%29%2F2%5C2+=+7.03219120813669
x%5B2%5D+=+%28-%283%29-sqrt%28+969+%29%29%2F2%5C2+=+-8.53219120813669

Quadratic expression 2x%5E2%2B3x%2B-120 can be factored:
2x%5E2%2B3x%2B-120+=+2%28x-7.03219120813669%29%2A%28x--8.53219120813669%29
Again, the answer is: 7.03219120813669, -8.53219120813669. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B3%2Ax%2B-120+%29

Hope this helps!