$\displaystyle \\\text{If }X\text{has a normal distribution with mean }1 \text{ and variance }4\text{, then }\\\\P[X^2-2X\leq 8]=?$

My question whether there is another way of approaching this problem other than realizing the probability in question can be rewritten as:

$\displaystyle \\P[X^2-2X\leq 8]= & P[X^2-2X+1\leq8+1]\\\\ & =P[(X-1)^2\leq 9]\\\\ & =P[-3\leq X-1\leq3]\\\\ & =P[-1.5\leq \frac{X-1}{2}\leq1.5]$

The reason that I ask is that I had a hard time seeing it.