Solve (2-n)^2+a^2+4+(3-a)^2=n^2+9

$$( 2 - n ) ^ { 2 } + a ^ { 2 } + 4 + ( 3 - a ) ^ { 2 } = n ^ { 2 } + 9$$

Solve for n

$$n=\frac{a^{2}-3a+4}{2}$$

Steps for Solving Linear Equation

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

$$4-4n+n^{2}+a^{2}+4+\left(3-a\right)^{2}=n^{2}+9$$

Add 4 and 4 to get 8.

$$8-4n+n^{2}+a^{2}+\left(3-a\right)^{2}=n^{2}+9$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3-a\right)^{2}.

$$8-4n+n^{2}+a^{2}+9-6a+a^{2}=n^{2}+9$$

Add 8 and 9 to get 17.

$$17-4n+n^{2}+a^{2}-6a+a^{2}=n^{2}+9$$

Combine a^{2} and a^{2} to get 2a^{2}.

$$17-4n+n^{2}+2a^{2}-6a=n^{2}+9$$

Subtract n^{2} from both sides.

$$17-4n+n^{2}+2a^{2}-6a-n^{2}=9$$

Combine n^{2} and -n^{2} to get 0.

$$17-4n+2a^{2}-6a=9$$

Subtract 17 from both sides.

$$-4n+2a^{2}-6a=9-17$$

Subtract 17 from 9 to get -8.

$$-4n+2a^{2}-6a=-8$$

Subtract 2a^{2} from both sides.

$$-4n-6a=-8-2a^{2}$$

Add 6a to both sides.

$$-4n=-8-2a^{2}+6a$$

The equation is in standard form.

$$-4n=-2a^{2}+6a-8$$

Divide both sides by -4.

$$\frac{-4n}{-4}=\frac{-2a^{2}+6a-8}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$n=\frac{-2a^{2}+6a-8}{-4}$$

Divide -8-2a^{2}+6a by -4.

$$n=\frac{a^{2}}{2}-\frac{3a}{2}+2$$

Solve for a (complex solution)

$$a=\frac{\sqrt{8n-7}+3}{2}$$

$$a=\frac{-\sqrt{8n-7}+3}{2}$$

Solve for a

$$a=\frac{\sqrt{8n-7}+3}{2}$$

$$a=\frac{-\sqrt{8n-7}+3}{2}\text{, }n\geq \frac{7}{8}$$