I would not say that "x= 2" was a solution to the original inequality. However, since this is a polynomial, so a continuous function, places where "p(x)> q(x)" and "p(x)< q(x)" are always separated by places where "p(x)= q(x)". Often the best way to solve an inequality is to first solve the associated equation.
Here, the inequality is \(\displaystyle (4-x^2)^2>(x^2-4x+4)^2\). The "associated equarion" is \(\displaystyle (4- x^2)^2= (x^2- 4x+ 4)^2\). That is the same as \(\displaystyle (2- x)^2(2+ x)^2= (x- 2)^4\). It is obvious that x= 2 is a solution. If x is not 2 we can divide both sides by \(\displaystyle (x- 2)^2\) to get \(\displaystyle (2+ x)^2= (2- x)^2\). That is the same as \(\displaystyle 4+ 4x+ x^2= x^2- 4x+ 4\) which reduces to \(\displaystyle 4x= -4x\) and then to \(\displaystyle x= 0\). That is, \(\displaystyle (4- x^2)^2= (x^2- 4x+ 4)^2\) at x= 0 and x= 2.
That means that the inequality may be satisfied in three intervals, \(\displaystyle -\infty< x< 0\), \(\displaystyle 0< x< 2\), and \(\displaystyle 2< x< \infty\). Now we just need to check the inequality at one x value in each of those intervals.
For \(\displaystyle x< 0\) take x= -1. \(\displaystyle (4- (-1)^2)^2= 3^2= 9\) and \(\displaystyle ((-1)^2- 4(-1)+ 4)^2= (1+ 4+ 4)^2= 9^2= 81\). 9< 81 so the inequality is false for all x< 0.
For \(\displaystyle 0< x< 2\) take x= 1. \(\displaystyle (4- 1^2)^2= 3^2= 9\) and \(\displaystyle ((1)^2- 4(1)+ 4)^2= 1^2= 1\). 9> 1 so the inequality is true for all 0< x< 2.
Finally, for \(\displaystyle x> 2\) take x= 3. \(\displaystyle (4- 3^2)^2= (-5)^2= 25\) and \(\displaystyle (3^2- 4(3)+ 4)^2= (9- 12+ 4)^2= 1\) 25> 1 so the inequality is true for all x>2.
\(\displaystyle (4-x^2)^2>(x^2-4x+4)^2\) is true for \(\displaystyle (0. 2)\cup (2, \infty)\), all positive x except 2.