Question 75666
{{{x^2-10x+24=0}}}
List the possible numbers that multiply to 24:
1,2,3,4,6,8,12,24
And you could also have negative numbers since (-8)(-3)=24. So of these possible factors, which 2 numbers add to -10? Lets look at (-8) and (-3), they multiply to 24 but add to -11, so that doesn't work. Lets look at  (-12) and (-2): they add to -14, that doesn't work either. Finally lets look at (-6) and (-4): they add to -10 and that works. In practice this might take longer since I already knew the answer but just wanted to show the process. So the polynomial {{{x^2-10x+24=0}}} factors to
{{{(x-6)(x-4)=0}}} which if foiled becomes {{{x^2-10x+24=0}}} again. So now set each factor equal to zero individually. The reason why this works is because we can say
{{{pq=0}}}
{{{p=0}}} divide both sides by q
and 
{{{q=0}}} divide both sides by p. So if either p or q is 0 (or both) then the entire equation equals zero.
The same applies to {{{(x-6)(x-4)=0}}}
{{{((x-6)*cross(x-4))/(cross(x-4))=0/(x-4)}}} divide both sides by (x-4)
{{{x-6=0}}}
{{{x+cross(-6+6)=0+6}}} Add 6 to both sides
{{{x=6}}}
{{{(cross(x-6)*(x-4))/(cross(x-6))=0/(x-6)}}} divide both sides by (x-6)
{{{x-4=0}}}
{{{x+cross(-4+4)=0+4}}} Add 4 to both sides
{{{x=4}}}
So the answers are x=6 or x=4


Or you could always plug the coefficients into the quadratic formula:
*[invoke quadratic "x", 1, -10, 24 ]