Solve 2/sqrt{5}=n2H-x/sqrt{left(2H-xright)^2+H^2}

$$\frac{ 2 }{ \sqrt{ 5 } } = n \frac{ 2H-x }{ \sqrt{ { \left(2H-x \right) }^{ 2 } + { H }^{ 2 } } }$$

Solve for n

$$n=\frac{2\sqrt{5\left(x^{2}-4Hx+5H^{2}\right)}}{5\left(2H-x\right)}$$

$$H\neq \frac{x}{2}\text{ and }\left(x\neq 0\text{ or }H\neq 0\right)$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{2}{\sqrt{5}} by multiplying numerator and denominator by \sqrt{5}.

$$\frac{2\sqrt{5}}{\left(\sqrt{5}\right)^{2}}=n\times \frac{2H-x}{\sqrt{\left(2H-x\right)^{2}+H^{2}}}$$

The square of \sqrt{5} is 5.

$$\frac{2\sqrt{5}}{5}=n\times \frac{2H-x}{\sqrt{\left(2H-x\right)^{2}+H^{2}}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2H-x\right)^{2}.

$$\frac{2\sqrt{5}}{5}=n\times \frac{2H-x}{\sqrt{4H^{2}-4Hx+x^{2}+H^{2}}}$$

Combine 4H^{2} and H^{2} to get 5H^{2}.

$$\frac{2\sqrt{5}}{5}=n\times \frac{2H-x}{\sqrt{5H^{2}-4Hx+x^{2}}}$$

Express n\times \frac{2H-x}{\sqrt{5H^{2}-4Hx+x^{2}}} as a single fraction.

$$\frac{2\sqrt{5}}{5}=\frac{n\left(2H-x\right)}{\sqrt{5H^{2}-4Hx+x^{2}}}$$

Use the distributive property to multiply n by 2H-x.

$$\frac{2\sqrt{5}}{5}=\frac{2nH-nx}{\sqrt{5H^{2}-4Hx+x^{2}}}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{2nH-nx}{\sqrt{5H^{2}-4Hx+x^{2}}}=\frac{2\sqrt{5}}{5}$$

Multiply both sides of the equation by 5.

$$5\left(5H^{2}-4Hx+x^{2}\right)^{-\frac{1}{2}}\left(2nH-nx\right)=2\sqrt{5}$$

Reorder the terms.

$$5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}\left(-nx+2Hn\right)=2\sqrt{5}$$

Use the distributive property to multiply 5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}} by -nx+2Hn.

$$-5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}nx+10\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}nH=2\sqrt{5}$$

Combine all terms containing n.

$$\left(-5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}x+10\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}H\right)n=2\sqrt{5}$$

The equation is in standard form.

$$\frac{10H-5x}{\sqrt{x^{2}-4Hx+5H^{2}}}n=2\sqrt{5}$$

Divide both sides by -5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}x+10\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}H.

$$\frac{\frac{10H-5x}{\sqrt{x^{2}-4Hx+5H^{2}}}n\sqrt{x^{2}-4Hx+5H^{2}}}{10H-5x}=\frac{2\sqrt{5}\sqrt{x^{2}-4Hx+5H^{2}}}{10H-5x}$$

Dividing by -5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}x+10\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}H undoes the multiplication by -5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}x+10\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}H.

$$n=\frac{2\sqrt{5}\sqrt{x^{2}-4Hx+5H^{2}}}{10H-5x}$$

Divide 2\sqrt{5} by -5\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}x+10\left(x^{2}-4Hx+5H^{2}\right)^{-\frac{1}{2}}H.

$$n=\frac{2\sqrt{5x^{2}-20Hx+25H^{2}}}{5\left(2H-x\right)}$$

Solve for H

$$\left\{\begin{matrix}H=\frac{x\left(-5n^{2}-\sqrt{5n^{2}-4}+4\right)}{10\left(1-n^{2}\right)}\text{, }&\left(n<1\text{ and }x<0\text{ and }n\geq \frac{2\sqrt{5}}{5}\right)\text{ or }\left(n>-1\text{ and }x>0\text{ and }n\leq -\frac{2\sqrt{5}}{5}\right)\text{ or }\left(n>1\text{ and }x>0\right)\text{ or }\left(n<-1\text{ and }x<0\right)\\H=\frac{x\left(-5n^{2}+\sqrt{5n^{2}-4}+4\right)}{10\left(1-n^{2}\right)}\text{, }&\left(n\neq -1\text{ and }n\leq -\frac{2\sqrt{5}}{5}\text{ and }x>0\right)\text{ or }\left(n\neq 1\text{ and }n\geq \frac{2\sqrt{5}}{5}\text{ and }x<0\right)\\H=\frac{x}{4}\text{, }&\left(n=-1\text{ and }x>0\right)\text{ or }\left(n=1\text{ and }x<0\right)\\H<0\text{, }&n=-1\text{ and }x=0\\H>0\text{, }&n=1\text{ and }x=0\end{matrix}\right.$$