like 6x^2+2x-5=0
I can find solutions but they are generally wrong like (2x-5)(6x+2)=0.
I think polynomial long-division might work, come to think of it.
Since the discriminant is irrational it will not factor with rational coefficients. Here is a way it can be done.
$\displaystyle \displaystyle 6x^2+2x-5=0 \iff 6\left[x^2+\frac{1}{3}x-\frac{5}{6}\right]=0$
Now complete the square to get
$\displaystyle \displaystyle 6\left[ \left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{1}{36}-\frac{5}{6}\right]=6\left[ \left(x+\frac{1}{6} \right)^2-\frac{31}{36}\right]$
Now we can write this as the difference of squares
$\displaystyle \displaystyle 6\left[ \left(x+\frac{1}{6} \right)^2-\left(\frac{\sqrt{31}}{6}\right)^2\right]=6\left[ \left(x+\frac{1}{6} -\frac{\sqrt{31}}{6}\right)\left(x+\frac{1}{6} +\frac{\sqrt{31}}{6}\right)\right]$