It is given that (3x+2)(2x-5)=ax^2+kx+n

(3x+2)(2x-5)=ax^2+kx+n

=> 6x^2 + 4x - 15x - 10 = ax^2 + kx + n

=> 6n^2 - 11x - 10 = ax^2 + kx + n

Equate the coefficients of x^2, x and the numeric coefficient. That gives a = 6, k = -11 and n = -10

a - n + k

=> 6 -(-10) - 11

=> 6 + 10 - 11

=> 5

**The value of a - n + k = 5**

3x+2)(2x-5)=ax^2+kx+n

What is the value of a-n+k?

Use FOIL.

(3x + 2)(2x - 5)

6x^2 - 15x + 4x - 10

6x^2 - 11x - 10

6x^2 + -11x + -10 = ax^2 + kx + n

a = 6

k = -11

n = -10

Now use substitution.

a - n + k

6 - -10 + -11 = 5

**Answer: 5**

All we need to do is to perform the multiplication from the left side and then to compare the results.

We'll multiply the brackets:

(3x+2)(2x-5)= 6x^2 - 15x + 4x - 10

We'll combine like terms:

(3x+2)(2x-5) = 6x^2 - 11x - 10

Comparing both sides, we'll get:

6x^2 - 11x - 10 = ax^2 + kx + n

a = 6 , k = -11 , n = -10

Now, we'll calculate a - n + k = 6 - (-10) + (-11)

a - n + k = 6 + 10 - 11

a - n + k = 5

**The requested value of a - n + k = 5.**