In the complex numbers, where \(i^2=-1\), \(\dfrac{2-i}{-3+i}\) = ?

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We need to multiply by the conjugate of the denominator to rationalize all square roots.

$$ \frac{2-i}{-3+i} $$ $$ \frac{2-i}{-3+i}\cdot \frac{-3-i}{-3-i} $$ $$ = \frac{(2-i)(-3-i)}{(-3+i)(-3-i)} $$ $$ = \frac{-6+3i-2i+i^2}{9-i^2} $$ $$ = \frac{-6+i+(-1)}{9-(-1)} $$ $$ = \frac{-7+i}{10} $$ $$ = \boxed{-\frac{7}{10}+\frac{1}{10}i} $$