Consider $x^{2}-25$. Rewrite $x^{2}-25$ as $x^{2}-5^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(x-5\right)\left(x+5\right)=0$$
To find equation solutions, solve $x-5=0$ and $x+5=0$.
$$x=5$$ $$x=-5$$
Steps by Finding Square Root
Add $25$ to both sides. Anything plus zero gives itself.
$$x^{2}=25$$
Take the square root of both sides of the equation.
$$x=5$$ $$x=-5$$
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: $ax^{2}+bx+c=0$.
$$x^{2}-25=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $-25$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$x=\frac{0±\sqrt{0^{2}-4\left(-25\right)}}{2}$$
Square $0$.
$$x=\frac{0±\sqrt{-4\left(-25\right)}}{2}$$
Multiply $-4$ times $-25$.
$$x=\frac{0±\sqrt{100}}{2}$$
Take the square root of $100$.
$$x=\frac{0±10}{2}$$
Now solve the equation $x=\frac{0±10}{2}$ when $±$ is plus. Divide $10$ by $2$.
$$x=5$$
Now solve the equation $x=\frac{0±10}{2}$ when $±$ is minus. Divide $-10$ by $2$.
