equations with radicals

[har]

Hi so this is a long one so Im going to show the major steps that I did.
I dont know how to show the radical sign so I am using * in its place.
*y-2*-*5y+1*=-3
*y-2*=-3+*5y+1*
y-2=10-6*5y+1*+5y
-4y-8/-6=-6*5y+1*/-6
16y^2+64y+64=5y+1
16y^2-116y+28=0
when I plug this into the quadratic formula
I get a big number that reduces down to 112 and 1/4,but when I check my answer they both come out wrong

 

[mmm4444bot]

*y-2*-*5y+1*=-3

*y-2*=-3+*5y+1*

y-2=10-6*5y+1*+5y

-4y-8/-6=-6*5y+1*/-6 This line is not correct.

-2 - 10 = -12; you typed -8 instead of -12

Also, I would not divide by -6, at this point. Square first, to avoid having to square fractional expression.



y - 2 = 10 - 6 sqrt(5y + 1) + 5y

-4y - 12 = -6 sqrt(5y + 1)

(-4y - 12)^2 = (-6)^2 * [sqrt(5y + 1)]^2

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