Consider $\left(x-3\right)\left(x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.
$$x^{2}+10x+16-\left(x^{2}-9\right)=0$$
To find the opposite of $x^{2}-9$, find the opposite of each term.
$$x^{2}+10x+16-x^{2}+9=0$$
Combine $x^{2}$ and $-x^{2}$ to get $0$.
$$10x+16+9=0$$
Add $16$ and $9$ to get $25$.
$$10x+25=0$$
Subtract $25$ from both sides. Anything subtracted from zero gives its negation.
$$10x=-25$$
Divide both sides by $10$.
$$x=\frac{-25}{10}$$
Reduce the fraction $\frac{-25}{10}$ to lowest terms by extracting and canceling out $5$.
