Solve for f
f=\frac{5x+4}{x^{2}}
x\neq 0
Solve for x (complex solution)
\left\{\begin{matrix}x=\frac{\sqrt{16f+25}+5}{2f}\text{; }x=\frac{-\sqrt{16f+25}+5}{2f}\text{, }&f\neq 0\\x=-\frac{4}{5}\text{, }&f=0\end{matrix}\right.
Solve for x
\left\{\begin{matrix}x=\frac{\sqrt{16f+25}+5}{2f}\text{; }x=\frac{-\sqrt{16f+25}+5}{2f}\text{, }&f\neq 0\text{ and }f\geq -\frac{25}{16}\\x=-\frac{4}{5}\text{, }&f=0\end{matrix}\right.
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fx^{2}-4=5x
Add 5x to both sides. Anything plus zero gives itself.
fx^{2}=5x+4
Add 4 to both sides.
x^{2}f=5x+4
The equation is in standard form.
\frac{x^{2}f}{x^{2}}=\frac{5x+4}{x^{2}}
Divide both sides by x^{2}.
f=\frac{5x+4}{x^{2}}
Dividing by x^{2} undoes the multiplication by x^{2}.
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