## Solving the Quadratic Equation: (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0

This article will guide you through the steps of solving the given quadratic equation.

### 1. Expand the Equation:

First, we need to expand the equation by multiplying out the brackets:

**(8-5x)(x+2) = 8x + 16 - 5x² - 10x = -5x² - 2x + 16****4(x-2)(x+1) = 4(x² - x - 2) = 4x² - 4x - 8****2(x-2)(x+2) = 2(x² - 4) = 2x² - 8**

Now we can substitute these back into the original equation:

**-5x² - 2x + 16 + 4x² - 4x - 8 + 2x² - 8 = 0**

### 2. Simplify the Equation:

Combine the like terms to simplify the equation:

**x² - 6x = 0**

### 3. Factor the Equation:

Factor out the common factor *x*:

**x(x - 6) = 0**

### 4. Solve for x:

For the product of two factors to equal zero, at least one of them must be zero. Therefore, we have two possible solutions:

**x = 0****x - 6 = 0 => x = 6**

### 5. Solution:

The solutions to the quadratic equation (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0 are **x = 0** and **x = 6**.