Question

the zeros of equation 8y^2-3y=0

Answer 1

I think it helped you

Answer 2

Answer:

8y^2-3y

y(8y-3)=0

y=0 ; 8y-3=0

y=3/8

alpha=0 & beta=3/8

alpha+beta=0+3/8= -(-3/8)= -b/a

alpha*beta=(0)(3/8)=0=c/a