Solve n+2=sqrt{a-n}

$$n + 2 = \sqrt { a - n }$$

Solve for a

$$a=\left(n+1\right)\left(n+4\right)$$

$$n+2\geq 0$$

Steps for Solving Radical Equations

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

$$\sqrt{a-n}=n+2$$

Square both sides of the equation.

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

Subtract -n from both sides of the equation.

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

Subtracting -n from itself leaves 0.

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

Subtract -n from \left(n+2\right)^{2}.

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

Solve for a (complex solution)

$$a=\left(n+1\right)\left(n+4\right)$$

$$n=-2\text{ or }arg(n+2)<\pi$$

Solve for n (complex solution)

$$\left\{\begin{matrix}n=\frac{\sqrt{4a+9}-5}{2}\text{, }&a=-2\text{ or }arg(\frac{\sqrt{4a+9}-1}{2})<\pi \\n=\frac{-\sqrt{4a+9}-5}{2}\text{, }&arg(\frac{-\sqrt{4a+9}-1}{2})<\pi \end{matrix}\right.$$

Solve for n

$$n=\frac{\sqrt{4a+9}-5}{2}$$

$$a\geq -2$$