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

Mar 27, 2015

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.

$15 {x}^{2} - 11 x - 3$ has only a few possibilities to consider for factoring: with integers
$\left(15 x \pm \text{ something")(x +- " something}\right)$ or
$\left(5 x \pm \text{ something")(3x +- " something}\right)$

The "somethings have to be $1 , 3$ in one order or the other. That's 4 possibilities. (Although it is clear that $\left(5 x \pm 1\right) \left(3 x \pm 3\right)$ 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). ${b}^{2} - 4 a c = {\left(- 11\right)}^{2} - 4 \left(15\right) \left(3\right) = 121 + 180 = 301$
The discriminant is positive so there are 2 real solutions, but it is not a perfect square, so the solutions are not rational.

$x = \frac{- \left(- 11\right) \pm \sqrt{{\left(- 11\right)}^{2} - 4 \left(15\right) \left(- 3\right)}}{2 \left(15\right)} = \frac{11 \pm \sqrt{301}}{30}$