Simple Polynomial Algebra question to complete the square $4y^2+32y = 0$
$4(y^2+8y+64-64) = 0$
$4(y+4)^2 = 64$
Is that correct?

$4y^2+32y = 0$
$4(y^2+8y+16-16) = 0$
$4(y+4)^2 = 64$
$(y+4)^2 = 16$
$y+4= \pm4$
$y_1=-8$, $y_2=0$
