This problem looks quite hairy, but thankfully the trinomial we've been given to work with can be factored fairly straightforwardly. #x^2-7x-18# factors into the product of binomials #(x+2)(x-9)#, so we can substitute that factored form into the denominator on the right side of the equation, obtaining
#1-5/(x+2)=7/((x+2)(x-9)#
This sets things up very conveniently for our next step, which is to multiply both sides by the binomial #x+2#. Keep in mind that we'll need to distribute this binomial both to the #1# and to the #-5/(x+2)#. Doing so, we have
#(x+2)-(5(x+2))/(x+2)=(7(x+2))/((x+2)(x-9))#
We can make a few cancellations of #x+2# terms on the left and right sides to obtain
#x+2-5=7/(x-9)#
Combining terms on the left side, we get
#x-3=7/(x-9)#
Next, to get the #x-9# term out of the denominator, we multiply both sides by it:
#(x-3)(x-9)=(7(x-9))/(x-9)#
The #x-9# terms on the right side cancel, leaving us with #7#, and expanding the expression on the left gives us the quadratic #x^2-12x+27#, which leaves the equation as
#x^2-12x+27=7#
If we want to solve for #x#, we'll need to make the right side of our equation #0# to make it easier for us to find the zeroes of the quadratic. We can do this by subtracting #7# from both sides, giving us
#x^2-12x+20=0#
As it turns out, this quadratic can be neatly factored into #(x-10)(x-2)#, making our equation
#(x-10)(x-2)=0#
and giving us a solution of #x=10, 2#
