SOLUTION: how do i solve this equation x2+2x+5=0 and express them in the form a +bi

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Question 268075: how do i solve this equation
x2+2x+5=0 and express them in the form a +bi
Answer by Alan3354(69443) 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%2B2x%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=%282%29%5E2-4%2A1%2A5=-16.

The discriminant -16 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 -16 is + or - sqrt%28+16%29+=+4.

The solution is x%5B12%5D+=+%28-2%2B-i%2Asqrt%28+-16+%29%29%2F2%5C1+=++%28-2%2B-i%2A4%29%2F2%5C1+, or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B5+%29

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x = -1 ± 2i