Consider $S^{2}-36$. Rewrite $S^{2}-36$ as $S^{2}-6^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(S-6\right)\left(S+6\right)=0$$
To find equation solutions, solve $S-6=0$ and $S+6=0$.
$$S=6$$ $$S=-6$$
Steps by Finding Square Root
Take the square root of both sides of the equation.
$$S=6$$ $$S=-6$$
Steps Using the Quadratic Formula
Subtract $36$ from both sides.
$$S^{2}-36=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-36$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$S=\frac{0±\sqrt{0^{2}-4\left(-36\right)}}{2}$$
Square $0$.
$$S=\frac{0±\sqrt{-4\left(-36\right)}}{2}$$
Multiply $-4$ times $-36$.
$$S=\frac{0±\sqrt{144}}{2}$$
Take the square root of $144$.
$$S=\frac{0±12}{2}$$
Now solve the equation $S=\frac{0±12}{2}$ when $±$ is plus. Divide $12$ by $2$.
$$S=6$$
Now solve the equation $S=\frac{0±12}{2}$ when $±$ is minus. Divide $-12$ by $2$.
