[tex]\bf \cfrac{g}{e^2}-r=m-y\implies \cfrac{g}{e^2}=m-y+r\implies g=e^2(m-y+r) \\\\\\ \cfrac{g}{m-y+r}=e^2\implies \pm\sqrt{\cfrac{g}{m-y+r}}=e[/tex]
Answer:
±[tex]\sqrt{\frac{g}{m-y+r}}[/tex]
Step-by-step explanation:
