Calcule: ײ-4×+3=0

No problema em questão, temos que:

\[\eqalign{ & {x^2} - 4x + 3 = 0 \cr & \cr & a = 1 \cr & b = - 4 \cr & c = 3 \cr & \cr & x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \cdot 1 \cdot 3} }}{{2 \cdot 1}} \cr & x = \dfrac{{4 \pm \sqrt {16 - 12} }}{2} \cr & x = \dfrac{{4 \pm \sqrt 4 }}{2} \cr & x = \dfrac{{4 \pm 2}}{2} \cr & \cr & \boxed{x' = \dfrac{{4 + 2}}{2} = \dfrac{6}{2} = 3} \cr & \boxed{x'' = \dfrac{{4 - 2}}{2} = \dfrac{2}{2} = 1} }\]