Which of the following equations has `2` as a root?

Options

A. `x^(2)-4x+5=0`
B. `x^(2)+3x-12=0`
C. `2x^(2)-7x+6=0`
D. `3x^(2)-6x-2=0`

Explanation

To determine which of the given equations has `2` as a root, we will substitute `x = 2` into each equation and check if the result equals `0`. If it does, then `2` is a root of that equation.

### Step-by-Step Solution:

1. **Equation 1: \( x^2 - 4x + 5 = 0 \)**
- Substitute \( x = 2 \):
\[ 2^2 - 4(2) + 5 = 4 - 8 + 5 = 1 \]
- Since \( 1 \neq 0 \), `2` is not a root of this equation.

2. **Equation 2: \( x^2 + 3x - 12 = 0 \)**
- Substitute \( x = 2 \):
\[ 2^2 + 3(2) - 12 = 4 + 6 - 12 = -2 \]
- Since \( -2 \neq 0 \), `2` is not a root of this equation.

3. **Equation 3: \( 2x^2 - 7x + 6 = 0 \)**
- Substitute \( x = 2 \):
\[ 2(2^2) - 7(2) + 6 = 2(4) - 14 + 6 = 8 - 14 + 6 = 0 \]
- Since \( 0 = 0 \), `2` is a root of this equation.

4. **Equation 4: \( 3x^2 - 6x - 2 = 0 \)**
- Substitute \( x = 2 \):
\[ 3(2^2) - 6(2) - 2 = 3(4) - 12 - 2 = 12 - 12 - 2 = -2 \]
- Since \( -2 \neq 0 \), `2` is not a root of this equation.

### Conclusion:
The only equation that has `2` as a root is:
\[ 2x^2 - 7x + 6 = 0 \]