SOLUTION: How can you use a quadratic formula to solve the following equations? 0 = x^2 + x - 20 0 = x^2 - 5x + 6

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Question 22863: How can you use a quadratic formula to solve the following equations?
0 = x^2 + x - 20
0 = x^2 - 5x + 6
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I'm going to assume you want to solve these equations separately;
i.e. you do not want to solve them as a system of two quadratic
equation.
If so you use the quadratic formula to solve the 2nd one as follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B6+=+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=%28-5%29%5E2-4%2A1%2A6=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+1+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%28-5%29-sqrt%28+1+%29%29%2F2%5C1+=+2

Quadratic expression 1x%5E2%2B-5x%2B6 can be factored:
1x%5E2%2B-5x%2B6+=+1%28x-3%29%2A%28x-2%29
Again, the answer is: 3, 2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B6+%29