Задача Задача №3648919
А
А
Answer: \(\boxed{x=2 \text{ or } x=3}\)
Key steps
Factor the quadratic:
\[ x^2-5x+6=(x-2)(x-3) \]
This works because \(-2\cdot(-3)=6\) and \(-2+(-3)=-5\).
Now use the zero-product rule:
\[ (x-2)(x-3)=0 \]
So either
\[ x-2=0 \quad\Rightarrow\quad x=2 \]
or
\[ x-3=0 \quad\Rightarrow\quad x=3 \]
Check
\[ 2^2-5\cdot2+6=4-10+6=0 \]
\[ 3^2-5\cdot3+6=9-15+6=0 \]
Both solutions are correct.