SOLUTION: solve -x²+3x=5 how do i solve this??

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Question 443454: solve -x²+3x=5
how do i solve this??
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
-x%5E2+%2B3x+=5
We don't like our coefficient to not be 1. So multiply both sides by -1.
x%5E2+-3x+=+-5
x%5E2+-3x+%2B5+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-3x%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=%28-3%29%5E2-4%2A1%2A5=-11.

The discriminant -11 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 -11 is + or - sqrt%28+11%29+=+3.3166247903554.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-3%2Ax%2B5+%29


So there are no solutions.