SOLUTION: (2/3)t^2 + (-4/3)t^2 -(1/5) = 0 please factor and use quadratic equation , explain, show work, this will be very helpful., Thank you so so much!

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Question 777090: (2/3)t^2 + (-4/3)t^2 -(1/5) = 0
please factor and use quadratic equation , explain, show work, this will be very helpful., Thank you so so much!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There may be a typo, but as written the problem does not provide a good example to demonstrate factoring, or using the quadratic formula.

To solve %282%2F3%29t%5E2+%2B+%28-4%2F3%29t%5E2+-%281%2F5%29+=+0+,
I would start by multiplying both sides of the equation times 15, so I do not have to deal with fractions.
15%2A%28%282%2F3%29t%5E2+%2B+%28-4%2F3%29t%5E2+-%281%2F5%29%29+=+15%2A0+
15%2A%282%2F3%29t%5E2+%2B+15%2A%28-4%2F3%29t%5E2+-15%2A%281%2F5%29%29+=+0+
10t%5E2-20t%5E2-3=0 --> -10t%5E2-3=0
Multiplying both sides by (-1), it turns into 10t%5E2%2B3=0
Now it looks easier, but it cannot be factored. Using the quadratic formula would be more trouble than it is worth, and it is obvious that it does not have any real solution.
Real numbers, have non-negative squares, so t%5E2%3E=0, 10t%5E2%3E=0, and
10t%5E2%2B3%3E=3 cannot be zero.
To look for complex solutions, I would first solve for t%5E2
10t%5E2%2B3=0-->10t%5E2=-3-->t%5E2=-3%2F10
Then the solutions would be