Consider $x^{2}-16$. Rewrite $x^{2}-16$ as $x^{2}-4^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(x-4\right)\left(x+4\right)=0$$
To find equation solutions, solve $x-4=0$ and $x+4=0$.
$$x=4$$ $$x=-4$$
Steps by Finding Square Root
Add $32$ to both sides. Anything plus zero gives itself.
$$2x^{2}=32$$
Divide both sides by $2$.
$$x^{2}=\frac{32}{2}$$
Divide $32$ by $2$ to get $16$.
$$x^{2}=16$$
Take the square root of both sides of the equation.
$$x=4$$ $$x=-4$$
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: $ax^{2}+bx+c=0$.
$$2x^{2}-32=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $0$ for $b$, and $-32$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
