SOLUTION: How would you solve this equation, x^2 + 2x - 80 = 0 ?

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Question 273969: How would you solve this equation, x^2 + 2x - 80 = 0 ?
Found 2 solutions by Tobiasz, richwmiller:
Answer by Tobiasz(54) About Me  (Show Source):
You can put this solution on YOUR website!
ok
x * x + 2x -80 = 0
2x + 2x - 80 = 0
4x - 80 = 0
4x = -80 Divide both by 4
x = -20

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-80+=+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%2A1%2A-80=324.

Discriminant d=324 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+324+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+324+%29%29%2F2%5C1+=+8
x%5B2%5D+=+%28-%282%29-sqrt%28+324+%29%29%2F2%5C1+=+-10

Quadratic expression 1x%5E2%2B2x%2B-80 can be factored:
1x%5E2%2B2x%2B-80+=+1%28x-8%29%2A%28x--10%29
Again, the answer is: 8, -10. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-80+%29