SOLUTION: y=-2x^(2)+2x+5

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Question 469985: y=-2x^(2)+2x+5
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
-2x2+2x+5=0
0=2x2-2x-5
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -2x%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%2A-2%2A5=44.

Discriminant d=44 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+44+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+44+%29%29%2F2%5C-2+=+-1.1583123951777
x%5B2%5D+=+%28-%282%29-sqrt%28+44+%29%29%2F2%5C-2+=+2.1583123951777

Quadratic expression -2x%5E2%2B2x%2B5 can be factored:
-2x%5E2%2B2x%2B5+=+-2%28x--1.1583123951777%29%2A%28x-2.1583123951777%29
Again, the answer is: -1.1583123951777, 2.1583123951777. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-2%2Ax%5E2%2B2%2Ax%2B5+%29
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