Solve the following quations:
`(2x+3)(3x-2)+2=0`

Answer

To solve the equation \((2x + 3)(3x - 2) + 2 = 0\), we will follow these steps:

### Step 1: Expand the equation
First, we need to expand the left-hand side of the equation:

\[ (2x + 3)(3x - 2) + 2 = 0 \]

Using the distributive property (also known as the FOIL method for binomials):

\[ = 2x \cdot 3x + 2x \cdot (-2) + 3 \cdot 3x + 3 \cdot (-2) + 2 \]

Calculating each term:

\[ = 6x^2 - 4x + 9x - 6 + 2 \]

### Step 2: Combine like terms
Now, combine the like terms:

\[ 6x^2 + (-4x + 9x) - 6 + 2 = 0 \]

This simplifies to:

\[ 6x^2 + 5x - 4 = 0 \]

### Step 3: Factor the quadratic equation
Next, we need to factor the quadratic equation \(6x^2 + 5x - 4 = 0\).

To factor, we look for two numbers that multiply to \(6 \times (-4) = -24\) and add to \(5\). These numbers are \(8\) and \(-3\).

So we can rewrite \(5x\) as \(8x - 3x\):

\[ 6x^2 + 8x - 3x - 4 = 0 \]

Now, group the terms:

\[ (6x^2 + 8x) + (-3x - 4) = 0 \]

Factoring out the common terms:

\[ 2x(3x + 4) - 1(3x + 4) = 0 \]

Now, factor out the common binomial:

\[ (2x - 1)(3x + 4) = 0 \]

### Step 4: Solve for \(x\)
Now we set each factor to zero:

1. \(2x - 1 = 0\)
\[ 2x = 1 \implies x = \frac{1}{2} \]

2. \(3x + 4 = 0\)
\[ 3x = -4 \implies x = -\frac{4}{3} \]

### Final Solution
The solutions to the equation \((2x + 3)(3x - 2) + 2 = 0\) are:

\[ x = \frac{1}{2} \quad \text{and} \quad x = -\frac{4}{3} \]

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