SOLUTION: solve 2x^2-3x=7 using quadratic function and rounding to the nearest tenth

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Question 613578: solve 2x^2-3x=7 using quadratic function and rounding to the nearest tenth
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-3x%2B-7+=+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=%28-3%29%5E2-4%2A2%2A-7=65.

Discriminant d=65 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--3%2B-sqrt%28+65+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-3%29%2Bsqrt%28+65+%29%29%2F2%5C2+=+2.76556443707464
x%5B2%5D+=+%28-%28-3%29-sqrt%28+65+%29%29%2F2%5C2+=+-1.26556443707464

Quadratic expression 2x%5E2%2B-3x%2B-7 can be factored:
2x%5E2%2B-3x%2B-7+=+%28x-2.76556443707464%29%2A%28x--1.26556443707464%29
Again, the answer is: 2.76556443707464, -1.26556443707464. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-3%2Ax%2B-7+%29