$\displaystyle x^2+7x+10=(x+2)(x+5)$

$\displaystyle x^2+5x+6=(x+2)(x+3)$

$\displaystyle x^2-8x+15=(x-3)(x-5)$

$\displaystyle x^2-x-30=(x-6)(x+5)$

These are ones I made up myself.

$\displaystyle x^2+x-2=(x+2)(x-1)$

$\displaystyle x^2-4-8=(x+2)(x-8)$

$\displaystyle x^2+3+9=(x+3)(x+3)$

$\displaystyle x^2+3-9=(x-3)(x-3)$

$\displaystyle x^2-4x+7$ This can not be solved right? As the only factor of 7 is 1 and 7, which I can not get -4 or 4 from? Are the others correct? And also does it matter if I have

$\displaystyle x^2-4-8=(x+2)(x-8)$ or $\displaystyle x^2-4-8=(x-8)(x+2)?$

Once a quadratic has be factorised, what is the next step? Can someone explain to be please.