Set $A=\{x: x$ is a prime number less than 4$\}$ $B=\left\{x: x^2-7 x-18=0, x \in R\right\}$ Then set $A$ and $B$ are

The set $A$ consists of prime numbers less than 4. The prime numbers less than 4 are 2 and 3. Thus, $A = \{2, 3\}$.

The set $B$ consists of the roots of the quadratic equation $x^2 - 7x - 18 = 0$. Factoring the equation, we get $(x - 9)(x + 2) = 0$. Thus, the roots are $x = 9$ and $x = -2$. So, $B = \{9, -2\}$.

Comparing the sets, $A = \{2, 3\}$ and $B = \{9, -2\}$. The number of elements in both sets is 2. Since the number of elements in $A$ is equal to the number of elements in $B$, the sets are equivalent.

Answer: Equivalent