Question

(x+2)^2 +11(x+2)+24 need solution set!!

Answer 1

is this equal to zero?

Answer 2

let u=x+2
so \[u^2=(u+2)^2\]
u^2+11u+24=0
(u+8)(u+3)=0
\[u=-8,u=-3\]
\[x+2=-8,x+2=-3\]
\[x=-8-2,x=-3-2\]
\[x=-10,x=-5\]

Answer 3

the solution set is {-10,-5}

Answer 4

Thank YOU

Answer 5

:)

Answer 6

0.2t=0.4+0.7t what is soulution set?

Answer 7

I hope you can help with this one please

Answer 8

0.2t-0.7t=0.4
-0.5t=0.4
t=0.4/(-0.5)
t=-4/5