First, note that we have x2 in the denominator. This tells us that x2 cannot equal to 0. So we have the boundary condition that x is not equal to 0. Now, to get rid of the denominator, we can multiply both sides by x2. This will give us the equation (x2+2)2=9x2. In order to solve this, we can move all terms to the left hand side and we will get (x2+2)2-9x2=0. Note that the expression 9x2 can be written as (3x)2. So we have (x2+2)2-(3x)2=0. This can be factorised as (x2+2+3x)(x2+2-3x)=0 using the product of sum and difference of two binomials. Note that the expression in the first brackets x2+3x+2 can also be written as x2+x+2x+2 which can be factorised as x(x+1)+2(x+1)=(x+1)(x+2). Similarly, the expression in the second brackets x2-3x+2 can be written as x2-x-2x+2=x(x-1)-2(x-1)=(x-1)(x-2). So by factorising the expressions in the brackets, we get for our equation (x+1)(x+2)(x-1)(x-2)=0. So the solutions for this are -2,-1, 1 and 2.