Question

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Answer 1

I know I'm supposed to use the -b+/- rt/b^2-4ac / 2(a) for the first one but I'm not sure how to set it up.

Answer 2

For the first equation,

\(3e^{2x} - 2e^x - 5 = 0\)

Replace e^x with y.

Then you will have a quadratic equation in y.
Solve it using the quadratic formula or factoring.
Then substitute back e^x for y.

Answer 3

So then it will be
3y^x - 2y - 5 = 0?

Answer 4

Close.
It'll be
3y^2 - 2y - 5 = 0

Answer 5

Alright thanks, that's probably where I was going wrong then!

Answer 6

Ok.

Answer 7

You're welcome.