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