Consider $t^{2}-25$. Rewrite $t^{2}-25$ as $t^{2}-5^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(t-5\right)\left(t+5\right)=0$$
To find equation solutions, solve $t-5=0$ and $t+5=0$.
$$t=5$$ $$t=-5$$
Steps by Finding Square Root
Add $5$ to both sides.
$$t^{2}=20+5$$
Add $20$ and $5$ to get $25$.
$$t^{2}=25$$
Take the square root of both sides of the equation.
$$t=5$$ $$t=-5$$
Steps Using the Quadratic Formula
Subtract $20$ from both sides.
$$t^{2}-5-20=0$$
Subtract $20$ from $-5$ to get $-25$.
$$t^{2}-25=0$$
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, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$t=\frac{0±\sqrt{0^{2}-4\left(-25\right)}}{2}$$
Square $0$.
$$t=\frac{0±\sqrt{-4\left(-25\right)}}{2}$$
Multiply $-4$ times $-25$.
$$t=\frac{0±\sqrt{100}}{2}$$
Take the square root of $100$.
$$t=\frac{0±10}{2}$$
Now solve the equation $t=\frac{0±10}{2}$ when $±$ is plus. Divide $10$ by $2$.
$$t=5$$
Now solve the equation $t=\frac{0±10}{2}$ when $±$ is minus. Divide $-10$ by $2$.
