Obviously, you're solving for e. Divide both sides by 9f to get
![e^(2)= (49)/(9f)](https://img.qammunity.org/2019/formulas/mathematics/high-school/sjv8fccayqtfbcnzprjepuvnzsdi87qzyo.png)
. Undo the square on the e by taking the square root of both sides, leaving you with
![e= ( √(49) )/( √(9f) )](https://img.qammunity.org/2019/formulas/mathematics/high-school/b43204jwck4emlrdvy1vx0ygn5zxe7u1te.png)
. Both 49 and 9 are perfect squares, so simplify accordingly to get that
![e=+/- (7)/(3 √(f) )](https://img.qammunity.org/2019/formulas/mathematics/high-school/j47n0hg2zhxauzcxi667gmy4kjl4epznk7.png)
. If you have to rationalize the denomiator, then your final answer would be
![e=+/- (7 √(f) )/(3f)](https://img.qammunity.org/2019/formulas/mathematics/high-school/6mk3ydr6zqbkbrlwrx1foq0ymxsxkuqn5v.png)