Solve the equation (y+3)(y−5) = 2y−2 for y.

The given equation is $(y+3)(y-5) = 2y-2$.
Expand the product on the left side using the FOIL method (First, Outer, Inner, Last).

$$(y+3)(y-5) = y \cdot y + y \cdot (-5) + 3 \cdot y + 3 \cdot (-5)$$ $$y^2 - 5y + 3y - 15$$ $$y^2 - 2y - 15$$

Subtract $2y$ from both sides:

$$y^2 - 2y - 2y - 15 = -2$$ $$y^2 - 4y - 15 = -2$$

Add 2 to both sides:

$$y^2 - 4y - 15 + 2 = 0$$ $$y^2 - 4y - 13 = 0$$

The quadratic formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
For the equation $y^2 - 4y - 13 = 0$, we have $a=1$, $b=-4$, and $c=-13$.
Substitute these values into the formula:

$$y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-13)}}{2(1)}$$ $$y = \frac{4 \pm \sqrt{16 - (-52)}}{2}$$ $$y = \frac{4 \pm \sqrt{16 + 52}}{2}$$ $$y = \frac{4 \pm \sqrt{68}}{2}$$

Simplify $\sqrt{68}$:

$$\sqrt{68} = \sqrt{4 \times 17} = \sqrt{4} \times \sqrt{17} = 2\sqrt{17}$$

Substitute this back into the expression for $y$:

$$y = \frac{4 \pm 2\sqrt{17}}{2}$$

Divide both terms in the numerator by 2:

$$y = \frac{4}{2} \pm \frac{2\sqrt{17}}{2}$$ $$y = 2 \pm \sqrt{17}$$

The solutions are $y = 2 + \sqrt{17}$ and $y = 2 - \sqrt{17}$.