To solve the quadratic equation \((3+i)w^2 - 2w + 3 - i = 0\), we use the quadratic formula:

\(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 3+i\), \(b = -2\), and \(c = 3-i\).

First, calculate the discriminant:

\(b^2 - 4ac = (-2)^2 - 4(3+i)(3-i)\)

\(= 4 - 4((3+i)(3-i))\)

\(= 4 - 4(9 - i^2)\)

\(= 4 - 4(9 + 1)\)

\(= 4 - 40\)

\(= -36\)

Now, substitute into the quadratic formula:

\(w = \frac{-(-2) \pm \sqrt{-36}}{2(3+i)}\)

\(= \frac{2 \pm 6i}{6+2i}\)

Multiply numerator and denominator by the conjugate of the denominator:

\(w = \frac{(2 \pm 6i)(6-2i)}{(6+2i)(6-2i)}\)

\(= \frac{12 \mp 4i + 36i \mp 12i^2}{36 + 4}\)

\(= \frac{12 \mp 4i + 36i \pm 12}{40}\)

\(= \frac{24 \pm 32i}{40}\)

\(= \frac{3}{5} \pm \frac{4}{5}i\)

Thus, the solutions are \(w = \frac{3}{5} + \frac{4}{5}i\) and \(w = -i\).