SOLUTION: Hi! Could you please help me with this problem? I have to find the solutions for 2x^2 + 4x + 7=0 using the quadratic equation...could you please help me? thank you so much!

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Question 378357: Hi! Could you please help me with this problem?
I have to find the solutions for 2x^2 + 4x + 7=0 using the quadratic equation...could you please help me?
thank you so much!

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Sub the coefficients for a, b & c
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B4x%2B7+=+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=%284%29%5E2-4%2A2%2A7=-40.

The discriminant -40 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -40 is + or - sqrt%28+40%29+=+6.32455532033676.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B4%2Ax%2B7+%29

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x = -1 ± i*sqrt(10)/2