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