Could someone help me with two problems involving radicals

6. solve √(14-x)+6=√(3-x)+9

7. Solve √(4y+20)-√(y-4)=6

I am having a really hard time I cam up with 2.889 for number six could someone please help me figure out what im doing wrong

√(14 - x) + 6 = √(3 - x) + 9
√(14 - x) - √(3 - x) = 3

Square both sides:

14 - x - 2√(14 - x)√(3 - x) + 3 - x = 9
√(14 - x)√(3 - x) = 4 - x
√[(14 - x)(3 - x)] = 4 - x

Square both sides again:

(14 - x)(3 - x) = x^2 - 8x + 16
x = 26 / 9

Note that 26 / 9 ≈ 2.889, so you were right.

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√(4y + 20) - √(y - 4) = 6

Square both sides:

4y + 20 - 2√(4y + 20)√(y - 4) + y - 4 = 36
√(4y + 20)√(y - 4) = (5/2)y - 10
√[(4y + 20)(y - 4)] = (5/2)[y - 4]

Square both sides again:

(4y + 20)(y - 4) = (25/4)[y - 4]^2
(25/4)[y - 4]^2 - (4y + 20)(y - 4) = 0
(y - 4)[(9/4)(y - 20)] = 0
(y - 4)(y - 20) = 0

y = 4 and y = 20

Done!