Solve a(n+2)^2=4a-n

$$a ( n + 2 ) ^ { 2 } = 4 a - n$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{1}{n+4}\text{, }&n\neq -4\\a\in \mathrm{R}\text{, }&n=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a\left(n^{2}+4n+4\right)=4a-n$$

Use the distributive property to multiply a by n^{2}+4n+4.

$$an^{2}+4an+4a=4a-n$$

Subtract 4a from both sides.

$$an^{2}+4an+4a-4a=-n$$

Combine 4a and -4a to get 0.

$$an^{2}+4an=-n$$

Combine all terms containing a.

$$\left(n^{2}+4n\right)a=-n$$

Divide both sides by n^{2}+4n.

$$\frac{\left(n^{2}+4n\right)a}{n^{2}+4n}=-\frac{n}{n^{2}+4n}$$

Dividing by n^{2}+4n undoes the multiplication by n^{2}+4n.

$$a=-\frac{n}{n^{2}+4n}$$

Divide -n by n^{2}+4n.

$$a=-\frac{1}{n+4}$$

Solve for n

$$\left\{\begin{matrix}\\n=0\text{, }&\text{unconditionally}\\n=-4-\frac{1}{a}\text{, }&a\neq 0\end{matrix}\right.$$