Solve A=(n+3)(n-3)-n^2+1

$$A = ( n + 3 ) ( n - 3 ) - n ^ { 2 } + 1$$

Solve for A

$$A=-8$$

Steps for Solving Linear Equation

Consider \left(n+3\right)\left(n-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$A=n^{2}-9-n^{2}+1$$

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

$$A=-9+1$$

Add -9 and 1 to get -8.

$$A=-8$$

Solve for n (complex solution)

$$n\in \mathrm{C}$$

$$A=-8$$

Solve for n

$$n\in \mathrm{R}$$

$$A=-8$$