Consider $\left(1+3x\right)\left(1-3x\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.
$$6+15x-9x^{2}\leq 1-\left(3x\right)^{2}$$
Expand $\left(3x\right)^{2}$.
$$6+15x-9x^{2}\leq 1-3^{2}x^{2}$$
Calculate $3$ to the power of $2$ and get $9$.
$$6+15x-9x^{2}\leq 1-9x^{2}$$
Add $9x^{2}$ to both sides.
$$6+15x-9x^{2}+9x^{2}\leq 1$$
Combine $-9x^{2}$ and $9x^{2}$ to get $0$.
$$6+15x\leq 1$$
Subtract $6$ from both sides.
$$15x\leq 1-6$$
Subtract $6$ from $1$ to get $-5$.
$$15x\leq -5$$
Divide both sides by $15$. Since $15$ is positive, the inequality direction remains the same.
$$x\leq \frac{-5}{15}$$
Reduce the fraction $\frac{-5}{15}$ to lowest terms by extracting and canceling out $5$.
