Obviously, you're solving for e. Divide both sides by 9f to get [tex] e^{2}= \frac{49}{9f} [/tex]. Undo the square on the e by taking the square root of both sides, leaving you with [tex]e= \frac{ \sqrt{49} }{ \sqrt{9f} } [/tex]. Both 49 and 9 are perfect squares, so simplify accordingly to get that [tex]e=+/- \frac{7}{3 \sqrt{f} } [/tex]. If you have to rationalize the denomiator, then your final answer would be [tex]e=+/- \frac{7 \sqrt{f} }{3f} [/tex]
