Solve 9=n-(2-k)^2

$$9 = n - ( 2 - k ) ^ { 2 }$$

Solve for n

$$n=\left(2-k\right)^{2}+9$$

Solution Steps

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

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

To find the opposite of 4-4k+k^{2}, find the opposite of each term.

$$9=n-4+4k-k^{2}$$

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

$$n-4+4k-k^{2}=9$$

Add 4 to both sides.

$$n+4k-k^{2}=9+4$$

Add 9 and 4 to get 13.

$$n+4k-k^{2}=13$$

Subtract 4k from both sides.

$$n-k^{2}=13-4k$$

Add k^{2} to both sides.

$$n=13-4k+k^{2}$$

Solve for k (complex solution)

$$k=\sqrt{n-9}+2$$

$$k=-\sqrt{n-9}+2$$

Solve for k

$$k=\sqrt{n-9}+2$$

$$k=-\sqrt{n-9}+2\text{, }n\geq 9$$