While it will not always be the case, start by assuming we are only dealing with integers and the factors of
#x^2+7x-18#
have the form
#(x+a)(x-b)# we know #a# and #-b# have different signs since their product is negative (#-18#)
We also know that #a# is greater than #b# since the coefficient of #x# is greater than zero.
There are only a limited number of possible integer values for #a# and #b# with #a>b# and #ab = 18#
#(a,b) = (18,1)# which would give #a-b = 17#; not what we want
#(a,b) = (9,2)# which would give #a-b = 7#; this matches the #x# coefficient of the given equation
#(a,b)= (6,3)# which would give #a-b=3#; not what we want
There are no other integer possibilities.
The factoring is
#(x+9)(x-2) = x^2 + 7x -18#
Since
#x^2 + 7x - 18 = 0#
either
#x+9 = 0# or #x-2=0#
So
either #x=-9# or #x=2#
