Consider $n^{2}-9$. Rewrite $n^{2}-9$ as $n^{2}-3^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(n-3\right)\left(n+3\right)=0$$
To find equation solutions, solve $n-3=0$ and $n+3=0$.
$$n=3$$ $$n=-3$$
Steps by Finding Square Root
Add $27$ to both sides. Anything plus zero gives itself.
$$3n^{2}=27$$
Divide both sides by $3$.
$$n^{2}=\frac{27}{3}$$
Divide $27$ by $3$ to get $9$.
$$n^{2}=9$$
Take the square root of both sides of the equation.
$$n=3$$ $$n=-3$$
Steps Using the Quadratic Formula
Quadratic equations like this one, with an $x^{2}$ term but no $x$ term, can still be solved using the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$, once they are put in standard form: $ax^{2}+bx+c=0$.
$$3n^{2}-27=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $0$ for $b$, and $-27$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
