SOLUTION: how do I using the quadratic formula to solve x^2-28=0

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Question 759370: how do I using the quadratic formula to solve x^2-28=0
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
how do I using the quadratic formula to solve x^2-28=0
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B-28+=+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=%280%29%5E2-4%2A1%2A-28=112.

Discriminant d=112 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+112+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%280%29%2Bsqrt%28+112+%29%29%2F2%5C1+=+5.29150262212918
x%5B2%5D+=+%28-%280%29-sqrt%28+112+%29%29%2F2%5C1+=+-5.29150262212918

Quadratic expression 1x%5E2%2B0x%2B-28 can be factored:
1x%5E2%2B0x%2B-28+=+%28x-5.29150262212918%29%2A%28x--5.29150262212918%29
Again, the answer is: 5.29150262212918, -5.29150262212918. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B-28+%29

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It's doing it the hard way.