SOLUTION: t-2(3-2t)=2t+9 3t-6-2t^2+4t=2t+9 7t-6-2t^2=2t+9 5t-6-2t^2=9 5t-2t^2=15 -2t^2+5t-15=0 I believe I have it right to this point, but I can not remember how to simplify when it i

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: t-2(3-2t)=2t+9 3t-6-2t^2+4t=2t+9 7t-6-2t^2=2t+9 5t-6-2t^2=9 5t-2t^2=15 -2t^2+5t-15=0 I believe I have it right to this point, but I can not remember how to simplify when it i      Log On


   

Question 337931: t-2(3-2t)=2t+9
3t-6-2t^2+4t=2t+9
7t-6-2t^2=2t+9
5t-6-2t^2=9
5t-2t^2=15
-2t^2+5t-15=0
I believe I have it right to this point, but I can not remember how to simplify when it is squared.
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
-2t^2+5t-15=0 This is correct. Well done!
This is a quadratic equation. The quadratic equation gives no real numbers as answers.
If you are familiar with imaginary numbers the answer is:
t=5/4 + sqrt(95)/4 * i, t=5/4 - sqrt(95)/4 * i (See below)
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Ed
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -2x%5E2%2B5x%2B-15+=+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=%285%29%5E2-4%2A-2%2A-15=-95.

The discriminant -95 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 -95 is + or - sqrt%28+95%29+=+9.74679434480896.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-2%2Ax%5E2%2B5%2Ax%2B-15+%29