Subtracting Sqrt?

Question

Sqrt 29w + 71 - Sqrt 5x + 51 = 2
Where w is a REAL number
I cannot seem to put this in a form
with the Sqrt's to calculate..

SOLVE FOR w

Answer 1

sqrt(29w + 71) - sqrt(5w + 51) = 2
sqrt(29w + 71) = 2 + sqrt(5w + 51)
square both sides
29w + 71 = (2 + sqrt(5w + 51))^2
29w + 71 = (2 + sqrt(5w + 51))(2 + sqrt(5w + 51))
29w + 71 = 4 + 2sqrt(5w + 51) + 2sqrt(5w + 51) + 5x + 51
29w + 71 = 4sqrt(5w + 51) + 5w + 55
24w + 16 = 4sqrt(5w + 51)
4(6w + 4) = 4sqrt(5w + 51)
6w + 4 = sqrt(5w + 51)
square both sides
(6w + 4)^2 = 5w + 51
(6w + 4)(6w + 4) = 5w + 51
36w^2 + 24w + 24w + 16 = 5w + 51
36w^2 + 48w + 16 = 5w + 51
36w^2 + 43w - 35 = 0

w = (-b ± sqrt(b^2 - 4ac))/(2a)

w = (-43 ± sqrt(43^2 - 4(36)(-35)))/(2(36))
w = (-43 ± sqrt(1849 + 5040))/72
w = (-43 ± sqrt(6889))/72
w = (-43 ± 83)/72
w = (-126/72) or (40/72)
w = (-7/4) or (5/9)

ANS : w = (-7/4) or (5/9)

Answer 2

you need to use parenthesis,
otherwise you can get totally wierd answers

anyway
i will just do whatever i understood:
Sqrt 29w + 71 - Sqrt 5x + 51 = 2
Sqrt 29w - Sqrt 5x = 2-71-51
sqrt(29w)=sqrt(5x) -120
so 29w = ( sqrt(5x) -120)^2
w = [( sqrt(5x) -120)^2]/29

Answer 3

Where does the square root extend?

Is it (√29 ) w + ..., or
(√(29w) ) + 71 - ..., or
(√(29w + 71) ) - ... or ... ?

Without any parentheses it means the first one.

Answer 4

there are two variables w and x
so to solve we need two equations

we can subtract only like terms
we cannot subtract a w termfrom an x term or vice versa