Question

3(1-y)^2+5(1-y)+2=0

Answer 1

can you solve
\[3u^2+5u+2=0\]

Answer 2

\[3u^2+2u+3u+2=0\]
\[u(3u+2)+1(3u+2)=0\]
\[(3u+2)(u+1)=0\]
\[3u+2=0 \text{ or } u+1=0\]

Answer 3

\[u=\frac{-2}{3} \text{ or } u=-1\]

Answer 4

what is different from my equation and your equation?

Answer 5

how about where you have 1-y I have u
right?

Answer 6

Mine is 3(1-y)^2 + 5(1-y)+2 = 0

Answer 7

so all you have to do now is solve both of these equations for y
1-y=-2/3 or 1-y=-1