Consider $y^{2}-1$. Rewrite $y^{2}-1$ as $y^{2}-1^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(y-1\right)\left(y+1\right)=0$$
To find equation solutions, solve $y-1=0$ and $y+1=0$.
$$y=1$$ $$y=-1$$
Steps by Finding Square Root
Add $9y^{2}$ to both sides.
$$3y^{2}+9y^{2}=12$$
Combine $3y^{2}$ and $9y^{2}$ to get $12y^{2}$.
$$12y^{2}=12$$
Divide both sides by $12$.
$$y^{2}=\frac{12}{12}$$
Divide $12$ by $12$ to get $1$.
$$y^{2}=1$$
Take the square root of both sides of the equation.
$$y=1$$ $$y=-1$$
Steps Using the Quadratic Formula
Add $9y^{2}$ to both sides.
$$3y^{2}+9y^{2}=12$$
Combine $3y^{2}$ and $9y^{2}$ to get $12y^{2}$.
$$12y^{2}=12$$
Subtract $12$ from both sides.
$$12y^{2}-12=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $12$ for $a$, $0$ for $b$, and $-12$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
