Solve Na^2=(20-a)^2-120

$$N a ^ { 2 } = ( 20 - a ) ^ { 2 } - 120$$

Solve for N

$$N=\frac{a^{2}-40a+280}{a^{2}}$$

$$a\neq 0$$

Steps for Solving Linear Equation

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(20-a\right)^{2}.

$$Na^{2}=400-40a+a^{2}-120$$

Subtract 120 from 400 to get 280.

$$Na^{2}=280-40a+a^{2}$$

The equation is in standard form.

$$a^{2}N=a^{2}-40a+280$$

Divide both sides by a^{2}.

$$\frac{a^{2}N}{a^{2}}=\frac{a^{2}-40a+280}{a^{2}}$$

Dividing by a^{2} undoes the multiplication by a^{2}.

$$N=\frac{a^{2}-40a+280}{a^{2}}$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{2\left(\sqrt{10\left(7N+3\right)}-10\right)}{N-1}\text{; }a=-\frac{2\sqrt{10}\left(\sqrt{7N+3}+\sqrt{10}\right)}{N-1}\text{, }&N\neq 1\\a=7\text{, }&N=1\end{matrix}\right.$$

Solve for a

$$\left\{\begin{matrix}a=\frac{2\left(\sqrt{10\left(7N+3\right)}-10\right)}{N-1}\text{; }a=-\frac{2\sqrt{10}\left(\sqrt{7N+3}+\sqrt{10}\right)}{N-1}\text{, }&N\neq 1\text{ and }N\geq -\frac{3}{7}\\a=7\text{, }&N=1\end{matrix}\right.$$