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
Subtract $3$ from both sides.
$$2x^{2}=21-3$$
Subtract $3$ from $21$ to get $18$.
$$2x^{2}=18$$
Divide both sides by $2$.
$$x^{2}=\frac{18}{2}$$
Divide $18$ by $2$ to get $9$.
$$x^{2}=9$$
Take the square root of both sides of the equation.
$$x=3$$ $$x=-3$$
Steps Using the Quadratic Formula
Subtract $21$ from both sides.
$$2x^{2}+3-21=0$$
Subtract $21$ from $3$ to get $-18$.
$$2x^{2}-18=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $0$ for $b$, and $-18$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
