(3x-2)(4x+3)-(x+3)(2x-5)=(2x+1)(5x-2)

(3x-2)(4x+3)-(x+3)(2x-5)=(2x+1)(5x-2)

$$\\\small{\text{
$
(3x-2)(4x+3) - (x+3)(2x-5) =(2x+1)(5x-2) \quad | \quad +(x+3)(2x-5)\\
$
}}$\\$
\small{\text{
$
(3x-2)(4x+3) =(2x+1)(5x-2) + (x+3)(2x-5)
$
}}$\\$
\small{\text{
$
12x^2 +9x -8x -6 = 10x^2 -4x +5x -2 + 2x^2 -5x +6x -15$
}}$\\$
\small{\text{
$
\underbrace{12x^2 -10x^2- 2x^2}_{0} \underbrace{+9x -8x +4x -5x +5x -6x}_{-x} \underbrace{-6 +2 +15}_{11} = 0 $
}}$\\\\$
\small{\text{
$
-x +11 = 0\\
$
}}$\\$
\small{\text{
$
x=11$
}}$$

General rules:

I: Left, outer, inner, right: (a + b)(c+ d) = ac + ad + bc + bd 

II:  a - b = -(b - a)

III: -(-w) = w

(3x - 2)(4x + 3) - (x + 3)(2x - 5) = (2x + 1)(5x - 2)        Use I:

12x² + 9x - 8x - 6 - (2x² - 5x +6x -15) = 10x²  - 4x + 5x - 2      Use III, add/subtract some x'es and numbers:

12x² + x  - 6 - 2x² + 5x - 6x + 15 = 10x² + x - 2         Add/subtract some x'es and numbers:

.....        ......         .....      .....                                     Add 2 to both sides:

.....        ......    

Can you take over from here?

BTW: It is very common to make mistakes based on a plus that should have been a minus or vica versa, especially when you have a minus before a paranthesis. When you have an expression like this, I'll highly recommend solving it this way:

-(x + 3)(2x - 5) = 

-(2x² - 5x +6x -15) =

- 2x² + 5x - 6x + 15 = ...

And NOT this way:

-(x + 3)(2x - 5) = 

- 2x² + 5x - 6x + 15 = ...

Both methods are mathematically correct, but the latter one makes it 1000 times more easy to make a mistake based on a wrong plus/minus.