The problem asks to multiply two complex numbers, $(2 + 5i)$ and $(3 - 2i)$.

1. Problem Description

The problem asks to multiply two complex numbers, $(2 + 5i)$ and $(3 - 2i)$.

2. Solution Steps

We need to multiply the two complex numbers using the distributive property (FOIL method).

$(a + bi)(c + di) = ac + adi + bci + bdi^2$

Recall that $i^2 = -1$.

First, multiply the first terms:

$2 * 3 = 6$

Second, multiply the outer terms:

$2 * (-2i) = -4i$

Third, multiply the inner terms:

$5i * 3 = 15i$

Fourth, multiply the last terms:

$5i * (-2i) = -10i^2$

So we have:

$6 - 4i + 15i - 10i^2$

Since $i^2 = -1$, we can replace $-10i^2$ with $-10 * (-1) = 10$.

Then the expression becomes:

$6 - 4i + 15i + 10$

Combine the real parts: $6 + 10 = 16$

Combine the imaginary parts: $-4i + 15i = 11i$

So we have: $16 + 11i$

3. Final Answer

$16 + 11i$