Answer: \(x=2\) or \(x=3\).
How to see it:
Factor the quadratic by finding two numbers that:
- multiply to the constant term \(6\),
- add to the middle coefficient \(-5\).
Those numbers are \(-2\) and \(-3\), because
\[ (-2)(-3)=6,\qquad (-2)+(-3)=-5. \]
So
\[ x^2-5x+6=(x-2)(x-3). \]
Set each factor equal to zero:
\[ x-2=0 \quad\Rightarrow\quad x=2, \] \[ x-3=0 \quad\Rightarrow\quad x=3. \]
Check:
\[ 2^2-5(2)+6=4-10+6=0, \] \[ 3^2-5(3)+6=9-15+6=0. \]
Both work, so the solutions are \(\boxed{x=2,\;3}\).