SOLUTION: HEY, how would you solve 3(p+1)^2=81?? and 5y^2+4=14

Question 165639: HEY, how would you solve 3(p+1)^2=81??
and 5y^2+4=14
Answer by NerdvanaGirl(8) (Show Source): You can put this solution on YOUR website!
For 3(p+1)^2=81
I would start by dividing both sides by 3
Therefore (p+1)^2=81/3 or (p+1)^2 = 27
I would then take the square root of each side
p+1 = radical 27
Then subtract 1
p=(radical 27) - 1 or p = -1 + (radical 27) or punch in radical 27 into a calculator and then subtract 1, which is 5.1961524 - 1 = approximately 4.196
For 5y^2+4=14
I would first subtract 4
so 5y^2=10
Then divide by 5
y^2=2
Take the square root
y= radical 2 or y= approximately 1.41421
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