SOLUTION: I am trying to solve y=x2-6x. so far column x is 0 1 2 3 4 column y is 0 -5 -8 a=1 b=-6 c=0. At school I understood where to get these #s now im lost. not sure how i need to do

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am trying to solve y=x2-6x. so far column x is 0 1 2 3 4 column y is 0 -5 -8 a=1 b=-6 c=0. At school I understood where to get these #s now im lost. not sure how i need to do      Log On


   

Question 168430: I am trying to solve y=x2-6x. so far column x is 0 1 2 3 4 column y is 0 -5 -8 a=1 b=-6 c=0. At school I understood where to get these #s now im lost. not sure how i need to do this
Answer by KnightOwlTutor(293) About Me  (Show Source):
You can put this solution on YOUR website!
This is a quadratic equation ax^2+bx+c=0
In your equation a=1 b=-6 and c=0
The y column is the value of y at the indicated x value.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%2B0+=+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=%28-6%29%5E2-4%2A1%2A0=36.

Discriminant d=36 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+36+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+36+%29%29%2F2%5C1+=+6
x%5B2%5D+=+%28-%28-6%29-sqrt%28+36+%29%29%2F2%5C1+=+0

Quadratic expression 1x%5E2%2B-6x%2B0 can be factored:
1x%5E2%2B-6x%2B0+=+1%28x-6%29%2A%28x-0%29
Again, the answer is: 6, 0. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B0+%29