If we start with #x^2-7x+10#, we can factor this by considering what *multiplys* to #10# (#1*10#) and *adds* to #-7#. I like to do this via table, like this:

#color(white)(.)#x#10##color(white)(.......)# **add to:**

..................................................

#1*10color(white)(00....)=10#

#-1*-10color(white)()=-10#

#1*-10color(white)(00)=-9#

#10*-1color(white)((.))=9#

#2*5color(white)(.......0.)=7#

#-2*-5color(white)(ct)=-7#

#2*-5color(white)(......)=3#

#5*-2color(white)(......)=3#

This might look like a mess (and I admit, it kinda is), but we now we know that the only factors of #x^2-7x+10# that multiply to 10 and add to #-7# are: #-2# and #-5#. That means that we can factor #x^2-7x+10# to #(x-2)(x-5)#. Now, to find the zeroes, we just set each parentheses equal to zero and solve for #x#, like this:

#x-2=0##color(white)(..........)##x-5=0#

#x=2##color(white)(................)##x=5#

Now we've got it. #x=2# and #5#. Nice job!