SOLUTION: Find the REAL Somution of the equation 2(x+1)^2+14(x+1)+20=0 The answers I have are A. (-5/2,-5) B. (-2,-6) C. (3/2,4) or D. (-5/2,-6) I keep getting -3,-6 what is the correc

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find the REAL Somution of the equation 2(x+1)^2+14(x+1)+20=0 The answers I have are A. (-5/2,-5) B. (-2,-6) C. (3/2,4) or D. (-5/2,-6) I keep getting -3,-6 what is the correc      Log On


   

Question 174538: Find the REAL Somution of the equation
2(x+1)^2+14(x+1)+20=0
The answers I have are
A. (-5/2,-5) B. (-2,-6) C. (3/2,4) or D. (-5/2,-6)
I keep getting -3,-6 what is the correct answer? I am stuck.

Can you help me? Thank you
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
2(x+1)^2+14(x+1)+20=0

Factor out the common factor of 2, or, better yet, divide both sides by 2:
(x+1)^2+7(x+1)+10=0

This factors into
((x+1) +5)*((x+1) + 2)=0
(x+6)(x+3)=0
x=-6 or x= -3

I don't see the correct answer here. Did you copy the multiple choices correctly??? Also, there are other methods to solve this, in case I confused you with this solution.

ALTERNATE SOLUTION (for anyone who did not understand the above solution!)
+%28x%2B1%29%5E2%2B7%28x%2B1%29%2B10=0
+x%5E2+%2B+2x+%2B+1+%2B+7x%2B7+%2B10+=0
x%5E2+%2B+9x+%2B+18=0

This trinomial factors into the product of two binomials:
+%28x%2B6%29%28x%2B3%29=0
x= -6 or x = -3

Different method, same solution!! I KNOW this is correct!! The given answers are WRONG!!

R^2