Question

Solve by factorisation method
$7y^2-32y+16=0$

Answer 1

$7y^2-32y+16=0$
$7y^2-28y-4y+16=0$
$7y(y-4)-4(y-4)=0$
$(y-4)(7y-4)=0$
$y=4=0$ or $7y-4=0$
$y=4$ or $7y=4$
i.e., $y=\frac{4}{7}$