Which equation has \(2\) as a root?
\(x^2-4x+5=0\)
\(x^2+3x-12=0\)
\(2x^2-7x+6=0\)
\(3x^2-6x-2=0\)
We are asked to check which equation has \(x = 2\) as a root. A number is a root of an equation if, after putting that number in place of \(x\), the left side becomes \(0\).
Option A: \(x^2 - 4x + 5\) Put \(x = 2\): \(2^2 - 4(2) + 5 = 4 - 8 + 5 = 1\). Since \(1 \neq 0\), this is not a root.
Option B: \(x^2 + 3x - 12\) Put \(x = 2\): \(2^2 + 3(2) - 12 = 4 + 6 - 12 = -2\). Since \(-2 \neq 0\), this is not a root.
Option C: \(2x^2 - 7x + 6\) Put \(x = 2\): \(2(2^2) - 7(2) + 6 = 2(4) - 14 + 6 = 8 - 14 + 6 = 0\). Since the value is exactly \(0\), \(x = 2\) is a root of this equation.
Option D: \(3x^2 - 6x - 2\) Put \(x = 2\): \(3(2^2) - 6(2) - 2 = 3(4) - 12 - 2 = 12 - 12 - 2 = -2\). Since \(-2 \neq 0\), this is not a root.
So, only Option C works. Therefore, the correct answer is C.