$X\sim N(2,3^2)$
How do you find $b$ where $P(-b<X<b)=0.95$ other than trial and error? You can't directly transform to $z$ because if you find an appropriate $z$, transforming back will give you the difference between $b$ and $z$, not $\pm z$.
I know it's a simple question but I'm stuck in a rut.
I don't think there's a purely analytical way to do this, but perhaps I can offer something a little more systematic than guess and check:
Lets first consider the much simpler problem of a mean $0$ normal RV with variance $9$.
In that case, we know from lookup tables that $b\approx 1.96\times 3$ will solve our problem. However, the shift upwards by two reduces the probability to:
$$P(-b<X<b)=\Phi\left(\frac{b-2}{3}\right)- \Phi\left(\frac{-b-2}{3}\right)<0.95$$
Now, lets augment this $b$ by another value $\delta>0$ so that:
$$P(-b<X<b)=\Phi\left(\frac{b-2+\delta}{3}\right)- \Phi\left(\frac{-b-2-\delta}{3}\right)=0.95$$
Therefore, you can simply solve (numerically), the equation above (I did it using Excel's Goal Seek) as a function of $\delta$ to get 0.95. I got 6.967891559 from Excel, similar to you. Unfortunately, the Gaussian CDF is not an easy function to work with analytically.