Question

How do you solve `15x^2-11x-3=0`?

Answer 1

I always solve quadratic equations the same way. Spend 2 to 3 minutes trying to factor using integers or square root method, then use the quadratics formula.

`15x^2-11x-3` has only a few possibilities to consider for factoring: with integers
`(15x+-" something")(x +- " something")` or
`(5x+-" something")(3x +- " something")`

The "somethings have to be `1, 3` in one order or the other. That's 4 possibilities. (Although it is clear that `(5x+-1)(3x+-3)` won't work, because the product of those two has every term divisible by `3`, and the question asked doesn't.
None of the other appear to work, so check the disciminant (some people do this first). `b^2-4ac=(-11)^2-4(15)(3)=121+180=301`
The discriminant is positive so there are 2 real solutions, but it is not a perfect square, so the solutions are not rational.

`x=(-(-11)+-sqrt((-11)^2-4(15)(-3)))/(2(15))=(11+-sqrt301)/30`