Solve for K
K=\frac{4\sqrt{5}x-5}{x^{2}}
x\neq 0
Solve for x (complex solution)
\left\{\begin{matrix}x=\frac{\sqrt{5}\left(\sqrt{4-K}+2\right)}{K}\text{; }x=\frac{\sqrt{5}\left(-\sqrt{4-K}+2\right)}{K}\text{, }&K\neq 0\\x=\frac{\sqrt{5}}{4}\text{, }&K=0\end{matrix}\right.
Solve for x
\left\{\begin{matrix}x=\frac{\sqrt{5}\left(\sqrt{4-K}+2\right)}{K}\text{; }x=\frac{\sqrt{5}\left(-\sqrt{4-K}+2\right)}{K}\text{, }&K\neq 0\text{ and }K\leq 4\\x=\frac{\sqrt{5}}{4}\text{, }&K=0\end{matrix}\right.
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Kx^{2}-4\sqrt{5}x=-5
Subtract 5 from both sides. Anything subtracted from zero gives its negation.
Kx^{2}=-5+4\sqrt{5}x
Add 4\sqrt{5}x to both sides.
x^{2}K=4\sqrt{5}x-5
The equation is in standard form.
\frac{x^{2}K}{x^{2}}=\frac{4\sqrt{5}x-5}{x^{2}}
Divide both sides by x^{2}.
K=\frac{4\sqrt{5}x-5}{x^{2}}
Dividing by x^{2} undoes the multiplication by x^{2}.
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