` 2x^(2) - 9x+10 = 0`
` y^(2) - 18y+72 = 0`

Options

A. `x gt y`
B. ` x lt y`
C. ` x ge y`
D. ` x le y`

Explanation

To solve the equations \(2x^2 - 9x + 10 = 0\) and \(y^2 - 18y + 72 = 0\) and establish a relationship between \(x\) and \(y\), we will follow these steps:

### Step 1: Solve the equation for \(x\)
We start with the equation:
\[ 2x^2 - 9x + 10 = 0 \]
To factor this quadratic equation, we look for two numbers that multiply to \(2 \times 10 = 20\) and add up to \(-9\). The numbers \(-4\) and \(-5\) fit this requirement.

We can rewrite the equation as:
\[ 2x^2 - 4x - 5x + 10 = 0 \]
Now, we group the terms:
\[ (2x^2 - 4x) + (-5x + 10) = 0 \]
Factoring out the common terms gives us:
\[ 2x(x - 2) - 5(x - 2) = 0 \]
Now, we can factor out \((x - 2)\):
\[ (2x - 5)(x - 2) = 0 \]
Setting each factor to zero gives us the solutions for \(x\):
\[ 2x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{2} \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Step 2: Solve the equation for \(y\)
Next, we solve the equation:
\[ y^2 - 18y + 72 = 0 \]
We need to find two numbers that multiply to \(72\) and add up to \(-18\). The numbers \(-6\) and \(-12\) work.

We can rewrite the equation as:
\[ y^2 - 6y - 12y + 72 = 0 \]
Grouping the terms gives us:
\[ (y^2 - 6y) + (-12y + 72) = 0 \]
Factoring out the common terms gives:
\[ y(y - 6) - 12(y - 6) = 0 \]
Factoring out \((y - 6)\):
\[ (y - 12)(y - 6) = 0 \]
Setting each factor to zero gives us the solutions for \(y\):
\[ y - 12 = 0 \quad \Rightarrow \quad y = 12 \]
\[ y - 6 = 0 \quad \Rightarrow \quad y = 6 \]

### Step 3: Establish the relationship between \(x\) and \(y\)
Now we have the values for \(x\) and \(y\):
- \(x = \frac{5}{2}\) or \(x = 2\)
- \(y = 12\) or \(y = 6\)

To establish a relationship, we compare the values:
- For \(x = 2\): \(y = 6\) (which means \(y > x\))
- For \(x = \frac{5}{2}\): \(y = 12\) (which also means \(y > x\))

Thus, in both cases, we find that \(y > x\).

### Final Conclusion
The relationship established is:
\[ y > x \]