Consider $m^{2}-9$. Rewrite $m^{2}-9$ as $m^{2}-3^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(m-3\right)\left(m+3\right)=0$$
To find equation solutions, solve $m-3=0$ and $m+3=0$.
$$m=3$$ $$m=-3$$
Steps by Finding Square Root
Take the square root of both sides of the equation.
$$m=3$$ $$m=-3$$
Steps Using the Quadratic Formula
Subtract $9$ from both sides.
$$m^{2}-9=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-9$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$m=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$
Square $0$.
$$m=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$
Multiply $-4$ times $-9$.
$$m=\frac{0±\sqrt{36}}{2}$$
Take the square root of $36$.
$$m=\frac{0±6}{2}$$
Now solve the equation $m=\frac{0±6}{2}$ when $±$ is plus. Divide $6$ by $2$.
$$m=3$$
Now solve the equation $m=\frac{0±6}{2}$ when $±$ is minus. Divide $-6$ by $2$.
