Consider $\left(x+1\right)\left(x-1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.
$$x^{2}-1=24$$
Add $1$ to both sides.
$$x^{2}=24+1$$
Add $24$ and $1$ to get $25$.
$$x^{2}=25$$
Take the square root of both sides of the equation.
$$x=5$$ $$x=-5$$
Steps Using the Quadratic Formula
Consider $\left(x+1\right)\left(x-1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.
$$x^{2}-1=24$$
Subtract $24$ from both sides.
$$x^{2}-1-24=0$$
Subtract $24$ from $-1$ to get $-25$.
$$x^{2}-25=0$$
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 $2$.
