Consider $x^{2}-9$. Rewrite $x^{2}-9$ as $x^{2}-3^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(x-3\right)\left(x+3\right)=0$$
To find equation solutions, solve $x-3=0$ and $x+3=0$.
$$x=3$$ $$x=-3$$
Steps by Finding Square Root
Add $9$ to both sides. Anything plus zero gives itself.
$$x^{2}=9$$
Take the square root of both sides of the equation.
$$x=3$$ $$x=-3$$
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: $ax^{2}+bx+c=0$.
$$x^{2}-9=0$$
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, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$
Square $0$.
$$x=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$
Multiply $-4$ times $-9$.
$$x=\frac{0±\sqrt{36}}{2}$$
Take the square root of $36$.
$$x=\frac{0±6}{2}$$
Now solve the equation $x=\frac{0±6}{2}$ when $±$ is plus. Divide $6$ by $2$.
$$x=3$$
Now solve the equation $x=\frac{0±6}{2}$ when $±$ is minus. Divide $-6$ by $2$.
