I. ` 3p^(2) + 17p+10 = 0`
II.` 10q^(2) +9q+2=0`

Options

A. if p is greater than q.
B. if p is smaller than q.
C. if p is equal q.
D. if p is either equal to or greater than q.

Explanation

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve the first equation for \( p \)
The first equation is:
\[ 3p^2 + 17p + 10 = 0 \]

To factor this quadratic equation, we look for two numbers that multiply to \( 3 \times 10 = 30 \) and add up to \( 17 \). The numbers \( 15 \) and \( 2 \) fit this requirement.

We can rewrite the equation as:
\[ 3p^2 + 15p + 2p + 10 = 0 \]

Now, we group the terms:
\[ (3p^2 + 15p) + (2p + 10) = 0 \]

Factoring out the common terms gives us:
\[ 3p(p + 5) + 2(p + 5) = 0 \]

Now, we can factor out \( (p + 5) \):
\[ (3p + 2)(p + 5) = 0 \]

Setting each factor to zero gives us:
1. \( 3p + 2 = 0 \) which leads to \( p = -\frac{2}{3} \)
2. \( p + 5 = 0 \) which leads to \( p = -5 \)

### Step 2: Solve the second equation for \( q \)
The second equation is:
\[ 10q^2 + 9q + 2 = 0 \]

We look for two numbers that multiply to \( 10 \times 2 = 20 \) and add up to \( 9 \). The numbers \( 5 \) and \( 4 \) fit this requirement.

We can rewrite the equation as:
\[ 10q^2 + 5q + 4q + 2 = 0 \]

Now, we group the terms:
\[ (10q^2 + 5q) + (4q + 2) = 0 \]

Factoring out the common terms gives us:
\[ 5q(2q + 1) + 2(2q + 1) = 0 \]

Now, we can factor out \( (2q + 1) \):
\[ (5q + 2)(2q + 1) = 0 \]

Setting each factor to zero gives us:
1. \( 5q + 2 = 0 \) which leads to \( q = -\frac{2}{5} \)
2. \( 2q + 1 = 0 \) which leads to \( q = -\frac{1}{2} \)

### Step 3: Compare the values of \( p \) and \( q \)
Now we have the values:
- For \( p \): \( -\frac{2}{3} \) and \( -5 \)
- For \( q \): \( -\frac{2}{5} \) and \( -\frac{1}{2} \)

### Step 4: Determine the order of \( p \) and \( q \)
We will compare the values of \( p \) and \( q \):
- The values of \( p \) are \( -5 \) and \( -\frac{2}{3} \).
- The values of \( q \) are \( -\frac{2}{5} \) and \( -\frac{1}{2} \).

To compare:
- \( -5 < -\frac{2}{3} \)
- \( -\frac{2}{5} > -\frac{1}{2} \)

Now, comparing \( -\frac{2}{3} \) and \( -\frac{2}{5} \):
- Convert to a common denominator (15):
- \( -\frac{2}{3} = -\frac{10}{15} \)
- \( -\frac{2}{5} = -\frac{6}{15} \)

Thus, \( -\frac{10}{15} < -\frac{6}{15} \) implies \( -\frac{2}{3} < -\frac{2}{5} \).

### Conclusion
From the comparisons:
- The largest value is \( -\frac{2}{5} \) (for \( q \)), followed by \( -\frac{1}{2} \) (for \( q \)), then \( -\frac{2}{3} \) (for \( p \)), and the smallest is \( -5 \) (for \( p \)).
- Therefore, we conclude that \( q > p \).

### Final Answer
Thus, the relation is:
\[ q > p \]

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