Answer
$\dfrac{11-8i}{5}$
Work Step by Step
Multiplying the numerator and the denominator by the conjugate of the denominator, the given expression, $
\dfrac{6-i}{2+i}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{6-i}{2+i}\cdot\dfrac{2-i}{2-i}
\\=
\dfrac{(6-i)(2-i)}{(2+i)(2-i)}
\\=
\dfrac{6(2)+6(-i)-i(2)-i(-i)}{(2)^2-(i)^2}
\\=
\dfrac{12-6i-2i+i^2}{4-i^2}
\\=
\dfrac{12-8i+i^2}{4-i^2}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{12-8i+(-1)}{4-(-1)}
\\=
\dfrac{12-8i-1}{4+1}
\\=
\dfrac{11-8i}{5}
.\end{array}