Consider $n^{2}-1$. Rewrite $n^{2}-1$ as $n^{2}-1^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(n-1\right)\left(n+1\right)=0$$
To find equation solutions, solve $n-1=0$ and $n+1=0$.
$$n=1$$ $$n=-1$$
Steps by Finding Square Root
Subtract $9$ from both sides.
$$n^{2}=10-9$$
Subtract $9$ from $10$ to get $1$.
$$n^{2}=1$$
Take the square root of both sides of the equation.
$$n=1$$ $$n=-1$$
Steps Using the Quadratic Formula
Subtract $10$ from both sides.
$$n^{2}+9-10=0$$
Subtract $10$ from $9$ to get $-1$.
$$n^{2}-1=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$n=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2}$$
Square $0$.
$$n=\frac{0±\sqrt{-4\left(-1\right)}}{2}$$
Multiply $-4$ times $-1$.
$$n=\frac{0±\sqrt{4}}{2}$$
Take the square root of $4$.
$$n=\frac{0±2}{2}$$
Now solve the equation $n=\frac{0±2}{2}$ when $±$ is plus. Divide $2$ by $2$.
$$n=1$$
Now solve the equation $n=\frac{0±2}{2}$ when $±$ is minus. Divide $-2$ by $2$.
