SOLUTION: 6x^2+17x+5=0

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Question 555494: 6x^2+17x+5=0
Answer by Maths68(1474) About Me  (Show Source):
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6x^2+17x+5=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 6x%5E2%2B17x%2B5+=+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=%2817%29%5E2-4%2A6%2A5=169.

Discriminant d=169 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-17%2B-sqrt%28+169+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2817%29%2Bsqrt%28+169+%29%29%2F2%5C6+=+-0.333333333333333
x%5B2%5D+=+%28-%2817%29-sqrt%28+169+%29%29%2F2%5C6+=+-2.5

Quadratic expression 6x%5E2%2B17x%2B5 can be factored:
6x%5E2%2B17x%2B5+=+6%28x--0.333333333333333%29%2A%28x--2.5%29
Again, the answer is: -0.333333333333333, -2.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+6%2Ax%5E2%2B17%2Ax%2B5+%29