Consider $t^{2}-4$. Rewrite $t^{2}-4$ as $t^{2}-2^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(t-2\right)\left(t+2\right)=0$$
To find equation solutions, solve $t-2=0$ and $t+2=0$.
$$t=2$$ $$t=-2$$
Steps by Finding Square Root
Divide both sides by $3$.
$$t^{2}=\frac{12}{3}$$
Divide $12$ by $3$ to get $4$.
$$t^{2}=4$$
Take the square root of both sides of the equation.
$$t=2$$ $$t=-2$$
Steps Using the Quadratic Formula
Divide both sides by $3$.
$$t^{2}=\frac{12}{3}$$
Divide $12$ by $3$ to get $4$.
$$t^{2}=4$$
Subtract $4$ from both sides.
$$t^{2}-4=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-4$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$t=\frac{0±\sqrt{0^{2}-4\left(-4\right)}}{2}$$
Square $0$.
$$t=\frac{0±\sqrt{-4\left(-4\right)}}{2}$$
Multiply $-4$ times $-4$.
$$t=\frac{0±\sqrt{16}}{2}$$
Take the square root of $16$.
$$t=\frac{0±4}{2}$$
Now solve the equation $t=\frac{0±4}{2}$ when $±$ is plus. Divide $4$ by $2$.
$$t=2$$
Now solve the equation $t=\frac{0±4}{2}$ when $±$ is minus. Divide $-4$ by $2$.
