# Given that $X$ is normal, find the probability that $(X-10)^2 <12$

Suppose that $X$ is a random variable that has a normal distribution with mean = 5 and standard deviation = 10. Evaluate the following probabilities:

$\mathrm{Prob}((X-10)^2 < 12)$

-
You are free to make changes like the one you discuss in your comment as long as they replace $(X-10)^2<12$ with something logically equivalent. (Changing means and standard deviations is irrelevant here, you are working with the distribution $X$.)
In this case, following your agenda, you would get $-\sqrt{12}<X-10<\sqrt{12}$, and then $-\sqrt{12}+10<X<\sqrt{12}+10$.
So, now I am trusting that you can compute this (normal) probability with the expression inside modified to this equivalent form: $P(-\sqrt{12}+10<X<\sqrt{12}+10)=$ ?