SOLUTION: Please help. use the quadratic formula to solve the equation.x^2-x=-5 I got: x=1+i√19/2 , 1-i√19/2 Am I correct?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Please help. use the quadratic formula to solve the equation.x^2-x=-5 I got: x=1+i√19/2 , 1-i√19/2 Am I correct?      Log On


   

Question 558358: Please help.
use the quadratic formula to solve the equation.x^2-x=-5
I got:
x=1+i√19/2 , 1-i√19/2
Am I correct?
Found 2 solutions by rfer, Alan3354:
Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
There are no zeros.
It opens up with the vertex at (.5,4.75)

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
x^2-x=-5
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%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=%28-1%29%5E2-4%2A1%2A5=-19.

The discriminant -19 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 -19 is + or - sqrt%28+19%29+=+4.35889894354067.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B5+%29

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You're correct.