Quick, your method certainly works, but as I pointed out to you before, I think you are doing too much work in using it.

In this case, we wish to factor $\displaystyle 4m^2 + 4m - 15$.

So solve:

$\displaystyle 4m^2 + 4m - 15 = 0$

$\displaystyle m = \frac{ -4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-15)}}{2 \cdot 4}$

$\displaystyle m = \frac{-4 \pm 16}{8} = \frac{3}{2}, \, -\frac{5}{2}$

So a quadratic polynomial in m with these roots is:

$\displaystyle \left ( m - \frac{3}{2} \right ) \left (m + \frac{5}{2} \right )$

Multiply each factor by 2 to clear the fractions:

$\displaystyle (2m - 3)(2m + 5)$

(The only thing you need to check is that the $\displaystyle m^2$ coefficient is correct. If not you will need to multiply by another factor to correct it.)

I suppose the main difference is a matter of style, but this method is more direct as it doesn't involve factoring at the end of the problem. It is also (like your method) kind of putting the cart before the horse: both of these methods depend on the quadratic formula which is designed to solve quadratics. The only reason to factor a quadratic is if you either don't know the quadratic formula or are trying to avoid using it. If you are going to solve a quadratic by factoring and don't wish to use the quadratic formula using either of these methods is kind of silly.

-Dan