SOLUTION: Three numbers are so related that the second number is 2 more than the first, and the third is 4 more than the first. Find the numbers if their squares is 56 more than three times

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Question 1025266: Three numbers are so related that the second number is 2 more than the first, and the third is 4 more than the first. Find the numbers if their squares is 56 more than three times the square of the smallest.
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
Three numbers are so related that
a, b, c
the second number is 2 more than the first,
b = a+2
and the third is 4 more than the first.
c = a+4
Find the numbers if their squares is 56 more than three times the square of the smallest.
a^2 + b^2 + c^2 = 3(c^2) + 56
Replace b with a+2, and c with a+4
a^2 + (a+2)^2 + (a+4)^2 = 3(a+4)^2 + 56
FOIL
a^2 + a^2 + 4a + 4 + a^2 + 8a + 16 = 3(a^2 + 8a + 16) + 56
Combine like terms, distribute the 3
3a^2 + 12a + 20 = 3a^2 + 24a + 48 + 56
Combine a on the left
12a - 24a = 48 + 56 - 20
-12a = 84
a = 84/-12
a = -7
then
b = -7+2
b = -5
and
c = -7+4
c = -3
:
:
you confirm this for yourself in the original equation
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Three numbers are so related that the second number is 2 more than the first, and the third is 4 more than the first. Find the numbers if their squares is 56 more than three times the square of the smallest.
1st and smallest number = 

2nd and 2nd smallest number = 

3rd and largest number =  

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