SOLUTION: Solve using the quadratic formula 12t^2-5t=2

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Question 371412: Solve using the quadratic formula
12t^2-5t=2
Answer by neatmath(302) About Me  (Show Source):
You can put this solution on YOUR website!

Put the equation into standard form, ie

12t%5E2-5t-2=0

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation at%5E2%2Bbt%2Bc=0 (in our case 12t%5E2%2B-5t%2B-2+=+0) has the following solutons:

t%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%2A12%2A-2=121.

Discriminant d=121 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+121+%29%29%2F2%5Ca.

t%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+121+%29%29%2F2%5C12+=+0.666666666666667
t%5B2%5D+=+%28-%28-5%29-sqrt%28+121+%29%29%2F2%5C12+=+-0.25

Quadratic expression 12t%5E2%2B-5t%2B-2 can be factored:
12t%5E2%2B-5t%2B-2+=+12%28t-0.666666666666667%29%2A%28t--0.25%29
Again, the answer is: 0.666666666666667, -0.25. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+12%2Ax%5E2%2B-5%2Ax%2B-2+%29