In class my teacher showed us an alternative method for factorising quadratics which are more awkward (i.e. the $a$ in $ax^2+bx+c$ is greater than 1).
The method is:
1. Take your quadratic (e.g. $3x^2+12x-15=0$)
2.Taking the product of the last and first term, and the middle term (here $-45$ and $12$)
3. Find the numbers which multiply to $-45$ and add to $12$ (here $15$ and $-3$)
4. Use these numbers to form the equation $15x^2-3x+15x-3$
5. Factorise this to: $3x(5x-1)+3(5x-1)=0$
6. Finally rearrange this to: $(5x-1)(3x+3)=0$
7. This gives a result of $x$ as $\frac15$ and $-1$
How ever this is not write as if you plug the numbers into the original equation you don't get $0$. Where have I gone wrong (appologies if the answer is fairly obvious). I have done this method with all the questions and only the equation $3x^2-6x=0$ worked using this method.