College Algebra (6th Edition)

$y = 3 \frac{+}{}\sqrt{13}$
$$y^2 - 6y - 4 =0$$ Since $(\frac{b}{2a})^2$ is $(\frac{-6}{2})^2 = -3^2 = 9$ for this case, we can re-write the original equation as follows: $$y^2 - 6y + 9 - 9 - 4 = 0$$ $$(y-3)^2 -13 = 0$$ $$y-3 =\frac{+}{}\sqrt{13}$$ $$y = 3 \frac{+}{}\sqrt{13}$$