SOLUTION: Please help me solve this question 2x^2+8x+5=0

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Question 816818: Please help me solve this question 2x^2+8x+5=0
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
2/3 (x+2)^2 = 1
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B8x%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=%288%29%5E2-4%2A2%2A5=24.

Discriminant d=24 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-8%2B-sqrt%28+24+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%288%29%2Bsqrt%28+24+%29%29%2F2%5C2+=+-0.775255128608411
x%5B2%5D+=+%28-%288%29-sqrt%28+24+%29%29%2F2%5C2+=+-3.22474487139159

Quadratic expression 2x%5E2%2B8x%2B5 can be factored:
2x%5E2%2B8x%2B5+=+2%28x--0.775255128608411%29%2A%28x--3.22474487139159%29
Again, the answer is: -0.775255128608411, -3.22474487139159. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B8%2Ax%2B5+%29