Consider $\left(\sqrt{5-3x}-2\right)\left(\sqrt{5-3x}+2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
$$\left(\sqrt{5-3x}\right)^{2}-2^{2}\leq 4$$
Calculate $\sqrt{5-3x}$ to the power of $2$ and get $5-3x$.
$$5-3x-2^{2}\leq 4$$
Calculate $2$ to the power of $2$ and get $4$.
$$5-3x-4\leq 4$$
Subtract $4$ from $5$ to get $1$.
$$1-3x\leq 4$$
Subtract $1$ from both sides.
$$-3x\leq 4-1$$
Subtract $1$ from $4$ to get $3$.
$$-3x\leq 3$$
Divide both sides by $-3$. Since $-3$ is negative, the inequality direction is changed.
