\(x^2-x+2\)
\(=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+2\)
\(=x\left(x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(x-\dfrac{1}{2}\right)+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}\right)+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\)'
Vậy đa thức \(x^2-x+2\) vô nghiệm.