Question

Find zeroes of 4u+8u square

Answer 1

let collect common from this 4u(1+2) now,first zero is 4u=0 and 1+2=0 OK this is zeroes

Answer 2

4u+8$u^{2}$

8$u^{2}$ + 4u

Let,

8$u^{2}$ + 4u = 0

4u ( 2u + 1) = 0

u = 0 and u = -1/2