\(x=28\implies z=\dfrac{28-\mu}\sigma\implies \mu-\sigma=28\)
\(x=34\implies z=\dfrac{34-\mu}\sigma\implies \mu+\sigma z=34\)
We need the exact value of \(z\) to find a proper solution, but a general one can still be found. Subtracting the first equation from the second gives
\((\mu+\sigma z)-(\mu-\sigma)=34-28\implies (z+1)\sigma=6\implies \sigma=\dfrac6{z+1}\)
Plug this into either equation and you get
\(\mu-\dfrac6{z+1}=28\implies \mu=28+\dfrac6{z+1}\)
(It's guaranteed that \(z
eq-1\) in this case because that already corresponds to \(x=28\).)
