Solve for `p:2p(p-3)+5(p-2)=0`

Options

A. `2,2 1/2`
B. `2 1/2, -2`
C. `-2 1/2, 2`
D. `-2 1/2,-2`

Answer

To solve the equation \(2p(p-3) + 5(p-2) = 0\), we will follow these steps:

### Step 1: Expand the equation
First, we will expand the terms in the equation.

\[ 2p(p-3) + 5(p-2) = 0 \]

Expanding \(2p(p-3)\):
\[ 2p^2 - 6p \]

Expanding \(5(p-2)\):
\[ 5p - 10 \]

Now, combine these results:
\[ 2p^2 - 6p + 5p - 10 = 0 \]

### Step 2: Combine like terms
Now, we will combine the like terms in the equation:
\[ 2p^2 - 6p + 5p - 10 = 0 \]
This simplifies to:
\[ 2p^2 - p - 10 = 0 \]

### Step 3: Factor the quadratic equation
Next, we need to factor the quadratic equation \(2p^2 - p - 10 = 0\).

To factor, we look for two numbers that multiply to \(2 \times -10 = -20\) and add to \(-1\) (the coefficient of \(p\)). The numbers that satisfy this are \(4\) and \(-5\).

Rewriting the equation using these factors:
\[ 2p^2 + 4p - 5p - 10 = 0 \]

### Step 4: Group the terms
Now, we will group the terms:
\[ (2p^2 + 4p) + (-5p - 10) = 0 \]

Factoring out the common terms:
\[ 2p(p + 2) - 5(p + 2) = 0 \]

### Step 5: Factor by grouping
Now we can factor out the common binomial factor \((p + 2)\):
\[ (p + 2)(2p - 5) = 0 \]

### Step 6: Solve for \(p\)
Now, we set each factor to zero:
1. \(p + 2 = 0\) leads to \(p = -2\)
2. \(2p - 5 = 0\) leads to \(2p = 5\) or \(p = \frac{5}{2}\)

### Final Answer
Thus, the solutions for \(p\) are:
\[ p = -2 \quad \text{or} \quad p = \frac{5}{2} \]

We can also express \(\frac{5}{2}\) as a mixed fraction:
\[ p = -2 \quad \text{or} \quad 2 \frac{1}{2} \]