Solve the quadratic equation y2 − 4y − 13 = 0 for y.

Your work correctly simplified the equation to:

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

Here, $a=1$, $b=-4$, and $c=-13$.

The quadratic formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Substitute the values of $a$, $b$, and $c$:

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

Simplify $\sqrt{68}$:

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

Substitute this back into the equation for $y$:

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

Divide both terms in the numerator by 2:

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