SOLUTION: (3+ √x)^2 + 3(3+√x)-10=0 u = (3+√x) u^2= (3+√x)^2 u^2 + 3u - 10 = 0 x= [-b � √(b^2-4ac)] � 2a -3 � √(3^2-4(1)(-10)) � (2)(1) -3 � √

Question 313371: (3+ √x)^2 + 3(3+√x)-10=0
u = (3+√x)
u^2= (3+√x)^2
u^2 + 3u - 10 = 0
x= [-b � √(b^2-4ac)] � 2a
-3 � √(3^2-4(1)(-10)) � (2)(1)
-3 � √(9 + 40 )� 2
-3 � √(49 ) � 2
-3 + 7 = 4/2 = 2
-3 � 7 = -10/2 = -5
What did I do wrong?
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Nothing. It looks good.
But you're not done yet.
Remember you solved for u.
You should have u in the quadratic formula and not x.
Now find x using your original substitution.

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SOLUTION: (3+ &#8730;x)^2 + 3(3+&#8730;x)-10=0 u = (3+&#8730;x) u^2= (3+&#8730;x)^2 u^2 + 3u - 10 = 0 x= [-b � &#8730;(b^2-4ac)] � 2a -3 � &#8730;(3^2-4(1)(-10)) � (2)(1) -3 � &#8730;

SOLUTION: (3+ √x)^2 + 3(3+√x)-10=0 u = (3+√x) u^2= (3+√x)^2 u^2 + 3u - 10 = 0 x= [-b � √(b^2-4ac)] � 2a -3 � √(3^2-4(1)(-10)) � (2)(1) -3 � √