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