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-315x2−11x−3 has only a few possibilities to consider for factoring: with integers
(15x+-" something")(x +- " something")(15x± something)(x± something) or
(5x+-" something")(3x +- " something")(5x± something)(3x± something)
The "somethings have to be 1, 31,3 in one order or the other. That's 4 possibilities. (Although it is clear that (5x+-1)(3x+-3)(5x±1)(3x±3) won't work, because the product of those two has every term divisible by 33, 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=301b2−4ac=(−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)/30x=−(−11)±√(−11)2−4(15)(−3)2(15)=11±√30130
