Consider $y^{2}-49$. Rewrite $y^{2}-49$ as $y^{2}-7^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(y-7\right)\left(y+7\right)=0$$
To find equation solutions, solve $y-7=0$ and $y+7=0$.
$$y=7$$ $$y=-7$$
Steps by Finding Square Root
Add $49$ to both sides. Anything plus zero gives itself.
$$y^{2}=49$$
Take the square root of both sides of the equation.
$$y=7$$ $$y=-7$$
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: $ax^{2}+bx+c=0$.
$$y^{2}-49=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-49$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$y=\frac{0±\sqrt{0^{2}-4\left(-49\right)}}{2}$$
Square $0$.
$$y=\frac{0±\sqrt{-4\left(-49\right)}}{2}$$
Multiply $-4$ times $-49$.
$$y=\frac{0±\sqrt{196}}{2}$$
Take the square root of $196$.
$$y=\frac{0±14}{2}$$
Now solve the equation $y=\frac{0±14}{2}$ when $±$ is plus. Divide $14$ by $2$.
$$y=7$$
Now solve the equation $y=\frac{0±14}{2}$ when $±$ is minus. Divide $-14$ by $2$.
