Write the conjugate of the complex numbers
`(2+5i)(-4+6i)`

Answer

To find the conjugate of the complex number \((2 + 5i)(-4 + 6i)\), we will follow these steps:

### Step 1: Multiply the complex numbers
We start by multiplying the two complex numbers:
\[ z = (2 + 5i)(-4 + 6i) \]

Using the distributive property (also known as the FOIL method for binomials), we have:
\[ z = 2 \cdot -4 + 2 \cdot 6i + 5i \cdot -4 + 5i \cdot 6i \]

Calculating each term:
- \(2 \cdot -4 = -8\)
- \(2 \cdot 6i = 12i\)
- \(5i \cdot -4 = -20i\)
- \(5i \cdot 6i = 30i^2\)

Since \(i^2 = -1\), we can replace \(30i^2\) with \(30 \cdot -1 = -30\).

Now, combining all the terms:
\[ z = -8 + 12i - 20i - 30 \]

### Step 2: Combine like terms
Now, we combine the real parts and the imaginary parts:
\[ z = (-8 - 30) + (12i - 20i) = -38 - 8i \]

### Step 3: Write the conjugate
The conjugate of a complex number \(z = x + yi\) is given by \(z^* = x - yi\). Therefore, for our complex number:
\[ z^* = -38 + 8i \]

### Final Answer
Thus, the conjugate of the complex number \((2 + 5i)(-4 + 6i)\) is:
\[ \boxed{-38 + 8i} \]